1. cartesian product × = × i) a b {1,3,5} {2,3} {(1,2), (1 ... · winglish coaching centre,...

Winglish Coaching Centre, Puduvayal 1 X std Mathematics Made Easy

1. CARTESIAN PRODUCT

1. Let A = {1, 2, 3} and B = {a, b}. Write AXB

and BXA

(b,3)} (b,2), (b,1),(a,3), (a,2), {(a,1),

{1,2,3} b}{a, AB

b)}(3,a),(3,b),(2,a),(2,b),(1,a),{(1,

b}{a,{1,2,3} BA

=

×=×

=

×=×

2. Find A x B , A x A and B x A

(i) A = {2, −2, 3} and B = {1, −4}

(ii) A = B = {p, q}

(iii) A = {m, n} ; B = f

n)} (n, m) (n, n) (m, m) {(m,

n} {m, n} {m, AA

Ø. A x B B x A Ø, B Since

Ø B ;n} {m, A iii)

q)} (q, q) (p, p) (q, p) {(p, A x B

q)} (q, p) (q, q) (p, p) {(p,

q} {p, q} {p,A x A

q)} (q, p) (q, q) (p, p) {(p, B x A

q} {p, B and q} {p, A ii)

3)} 3)(-4, 2)(1,- 2)(-4,- 2)(1, 2)(-4, {(1, A x B

3)} 2)(3,- 2)(3, (3,

3) 2)(-2,- 2)(-2, 3)(-2, 2)(2,- 2)(2, {(2, A x A

4)}- 1)(3, 4)(3,- 1)(-2, 4)(-2,- 1)(2, {(2, B x A

4}- {1, B

3} 2,- {2, A i)

=

×=×

===

==

=

=

×=

=

==

=

=

=

=

=

3. Let A = {1, 2, 3} and B = {x | x is a prime

number less than 10}. Find A x B and B x A.

} 3) (7, 2) (7, 1) (7, 3) (5, 2) (5,

1) (5, 3) (3, 2) (3, 1) 3)(3, (2, 2) (2, 1) (2, { A x B

} 7) (3, 5) (3, 3) (3, 2) 7)(3, (2, 5) (2,

3) (2, 2) 7)(2, (1, 5) (1, 3) (1, 2) (1, { B x A

7} 5, 3, {2, B

10}.than less number prime a is x | {x B

3} 2, {1, A

=

=

=

=

=

4. If A = {1,3,5} and B = {2,3} then

(i) find A×B and B × A.

(ii) Is A×B=B×A? If not why?

(iii) Show that n(A×B) = n(B×A) = n(A)× n(B)

6. (B)n n(A) A)(Bn B)(An

6 32 (A)n (B)n

6 2 3 (B)n (A)n

6 A)(Bn B)(An

2. (B)n 3;n(A) iii)

AB BA Thus

),(),(),( (1,2) ii)

(3,5)} (3,3), (3,1), (2,5), (2,3), {(2,1),

{1,3,5} {2,3} AB

(5,3)} (5,2), (3,3), (3,2), (1,3), {(1,2),

{2,3} {1,3,5} BA i)

=×=×=×

=×=×

=×=×

=×=×

==

×≠×

≠≠

=

×=×

=

×=×

133112 and

5. If A×B = {(3,2), (3,4), (5,2), (5,4)} find A and B.

{2,4} B

B}.A of scoordinate second all of {set B

{3,5} A

} BA of scoordinate first all of {set A

(5,4)} (5,2), (3,4), {(3,2), BA

=

×=

=

×=

=×

6. If B × A = {(−2, 3),(−2, 4),(0, 3),(0, 4),(3, 3),(3,

4)} find A and B.

4}{3, A

A}B of scoordinate second all of {set A

3} 0, {-2, B

} AB of scoordinate first all of {set B

=

×=

=

×=

7. If A = {5, 6} , B = {4, 5, 6} , C = {5, 6, 7} , Show

that A × A = (B × B) n (C × C) .

R.H.S L.H.S

(2) (1)

(2)----- 6)} 5)(6, 6)(6, 5)(5, {(5, C) (C B) (B

7)} 6)(7, (7,

5) 7)(7, 6)(6, 5)(6, 7)(6, 6)(5, 5)(5, {(5,

7} 6, {5, 7} 6, {5, CC

6)} (6, 5) (6,

4)6)(6, 5)(5, 4)(5,6)(5, (4, 5) (4, 4){(4,

6} 5, {4, 6} 5, {4, BB R.H.S

-(1)--- 6) (6, 5) (6, 6) (5, 5) {(5,

6} {5, 6} {5, AA L.H.S

=

=

=×∩×

=

×=×

=

×=×

=

×=×

8. Given A = {1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and

D = {1, 3, 5}, check if (A n C) × (B n D) = (A ×

B)n(C × D) is true?


true. is statementgiven the Hence

(2) (1)

-(2)----- 3)(3,5)} {(3, D) (CB) (A

5)} 3)(4, 1)(4, 5)(4, 3)(3, 1)(3, {(3, D x C

5)} 3)(3, 2)(3, (3,

5) 3)(2, 2)(2, 5)(2, 3)(1, 2)(1, {(1, B x A

: R.H.S

(1)----- } 5) 3)(3, (3, { D)(B C) (A

5} {3, D B

{3} C A

:H.S L.

=

=×∩×

=

=

=∩×∩

=∩

=∩

9. Let A = {x ∈ N | 1< x < 4} , B = {x ∈W|0 ≤

x ≤ 2} and C = {x∈ N |x <3} . Verify that

i) A × (B ∪ C) = (A × B) ∪ (A × C)

ii) A × (B ∩ C) = (A × B) ∩ (A × C)

i) A × (B ∪ C) = (A × B) ∪ (A × C)

(2)(1)

..(2)3,1)(3,2)}2,2)(3,0)(2,0)(2,1)({()CA(B)(A

)})(3,1)(3,2{(2,1)(2,2CA

)})(3,0)(3,1{(2,0)(2,1BA

...(1)3,1)(3,2)}2,2)(3,0)(2,0)(2,1)({()(

},,{

=

=×∪×

=×

=×

=∪×

=∪

CBA

CB 210

ii) A × (B ∩ C) = (A × B) ∩ (A × C)

(2)(1)

}..(2)(2,1)(3,1){()CA(B)(A

)})(3,1)(3,2{(2,1)(2,2CA

)})(3,0)(3,1{(2,0)(2,1BA

}...(1)(2,1)(3,1){()(

}{

=

=×∩×

=×

=×

=∩×

=∩

CBA

CB 1

10. Let A = {x ∈ W | x < 2} , B = {x ∈|1 < x ≤ 4}

and C = {3, 5} . Verify that

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)

A = {x ∈ W| x < 2} ,

A = {0, 1}

B = {x ∈N|1 < x ≤ 4}

B = {2, 3, 4}

C = {3, 5} .

i) A × (B ∪ C) = (A × B) ∪ (A × C)

proved Hence

(2) (1)

-(2)- 5)} (1, 4)(1,

3) (1, 2) 5)(1, (0, 4)(0, 3) (0, 2) {(0, C) x (A B) x (A

5)} (1, 3) (1, 5) 3)(0, {(0, C) (A

4)}(1, 3) (1, 2) 4)(1,(0, 3) (0, 2) {(0, B) (A

:R.H.S

-(1)- 5)} (1, 4)(1, 3) (1,

2) 5)(1, (0, 4)(0, 3) (0, 2) {(0, C)(B A

5} 4,3, {2, C)(B

:L.H.S

=

=∪

=×

=×

=∪×

=∪

(ii) A × (B n C) = (A × B) n (A × C)

proved. Hence

(2) (1)

(2)--- 3)} (1, 3) {(0, C) (AB) (A

5)} (1, 3) (1, 5) 3)(0, {(0, C) (A

4)}(1, 3) (1, 2) 4)(1,(0, 3) (0, 2) {(0, B) (A

:R.H.S

(1)--- 3)} (1, 3) {(0, C)(B A

{3} C) (B

: L.H.S

=

=×∩×

=×

=×

=∩×

=∩

(iii) (A U B) × C = (A × C) U (B × C)

(2) (1)

(2)----- } 5) 3)(4, 5)(4, 3)(3, 5)(3, (2,

3) 5)(2, 3)(1, 5)(1, 3)(0, (0, { C) (BC) (A

5)} (4, 3) (4, 5) 3)(3, (3, 5) (2, 3) {(2, C) (B

4)}(1, 3) (1, 2) 4)(1,(0, 3) (0, 2) {(0, B) (A

(1)----- } 5) 3)(4, 5)(4, 3)(3, 5)(3, (2,

3) 5)(2, 3)(1, 5)(1, 3)(0, (0, { C B)(A

4}3, 2, 1, {0, BA

=

=×∪×

=×

=×

=×∪

=∪

11. A = Set of all natural numbers less than 8,

B = Set of all prime numbers less than 8,

C = Set of even prime number. Verify that

i) (A ∩ B) × C = (A × C) ∩ (B × C)

ii) A × (B −C) = (A × B) − (A × C)

A = {1, 2, 3, 4, 5, 6, 7}

B = {2, 3, 5, 7}

C = {2}

i) (A n B) × C = (A × C) n (B × C)


proved. Hence

(2) (1)

)2)}.....(2 (7, 2) (5, 2) (3, 2) {(2, C) (B C) (A

2)} 2)(7, 2)(5, 2)(3, {(2, C) (B

2)} 2)(7, (6,

2) (5, 2) 2)(4, 2)(3, 2)(2, {(1, C) (A

:R.H.S

....(1) 2)} (7, 2) (5, 2) (3, 2) {(2, C B)(A

7} 5, 3, {2, B) (A

: L.H.S

=

=×∩×

=×

=×

=×∩

=∩

ii) A × (B − C) = (A × B) − (A × C)

proved. Hence

(2) (1)

-(2)--- 7)} (7, 5) (7, 3) 7)(7, (6, 5) (6, 3) (6,

7) (5, 5) (5, 3) (5, 7) (4, 5) 3)(4, 7)(4, 5)(3, (3,

3) (3, 7) (2, 5) (2, 3) (2, 7) (1, 5) 3)(1, {(1, C) (A - B) (A

2)} 2)(7, 2)(6, (5, 2) 2)(4, 2)(3, 2)(2, {(1, C) (A

7) 5)(7, 3)(7, 2)(7, (7,

7) 5)(6, 3)(6, 2)(6, 7)(6, 5)(5, 3)(5, 2)(5, (5,

7) 5)(4, 3)(4, 2)(4, 7)(4, 5)(3, 3)(3, 2)(3, (3,

7) 5)(2, 3)(2, 2)(2, 7)(2, 5)(1, 3)(1, 2)(1, {(1, B) (A

-(1)--- 7)} (7, 5) (7,

3) 7)(7, (6, 5) (6, 3) (6, 7) (5, 5) (5, 3) (5,

7) (4, 5) (4, 3) (4, 7) 5)(3, (3, 3) (3,

7) (2, 5) (2, 3) (2, 7) (1, 5) (1, 3) {(1, C) - (B A

7} 5, {3, C) - (B

=

=××

=×

=×

=×

=

2. RELATIONS 1. Define a relation between heights of

corresponding students using i) pair of

heights and students ii) an arrow diagram

R = {(heights, students)}

i) pair of heights and students

R= { (4.5, S1) (4.5, S4) (4.7, S9) (4.9, S10) (5, S3) (5, S5)

(5, S8) (5.1, S6) (5.2, S2) (5.2, S7)}

ii) an arrow diagram

2. Let A = {3,4,7,8} and B = {1,7,10}. Which of

the sets are relations from A to B?

(i) R1 ={(3,7), (4,7), (7,10), (8,1)}

(ii) R2= {(3,1), (4,12)}

(iii)R3= (3,7)(4,10)(7,7)(7,8)(8,11)(8,7)(8,10)}

A× B = {(3,1)(3,7)(3,10)(4,1)(4,7)(4,10)

(7,1)(7,7)(7,10)(8,1) (8,7) (8,10)}

(i) R1 ⊆ A×B

Thus, R1 is a relation from A to B.

(ii) (4,12)∈ R2 but (4,12)∉ A×B

So, R2 is not a relation from A to B.

(iii) (7,8) ∈ R3 but (7,8) ∉ A×B

So, R3 is not a relation from A to B.

3. Let A = {1, 2, 3, 7} and B = {3, 0, –1, 7}, which of

the following are relation from A to B ?

(i) R1 = {(2, 1), (7, 1)}

(ii) R2 = {(–1,1)}

(iii) R3 = {(2, –1), (7, 7), (1, 3)}

(iv) R4= {(7,–1), (0,3), (3,3), (0,7)}

A x B = {(1,3)(1,0) (1, -1) (1,7)(2,3) (2,0) (2, -1)

(2,7)(3,3)(3,0) (3,-1)(3,7)(7,3)(7,0)(7, -1)(7, 7)}

i) (2, 1) ∈ R1 but (2, 1) ∉A x B.

Hence R1 is not a relation.

ii) (-1, 1) ∈ R2, but (1, -1) ∉ A×B

Hence R2 is not a relation.

iii) R3 ⊆A x B,

Hence it is a relation from A to B.

iv) (0,3), (3,3), (0,7) ∈ R4 but ∉ A x B

Hence it is not a relation.

4. Let A = {1,3,5,7}, B = {4,8} and R be the

relation defined as “is less than”.

i) Write R as a subset of A x B.

ii) Define a relation R from A to B

ii) Write in xRy form

7R8 5R8, 3R8, 3R4, 1R8,R Or

(7,8)} (5,8), (3,8), (3,4), (1,8), {(1,4), R

}(7,4)(7,8) (5,8) (5,4) (3,8) (3,4) (1,8) {(1,4) BA

=

=

=×

5. Let A = {1, 2, 3, 4,..., 45} and R be the relation

defined as “is square of ” on A.


i) Write R as a subset of A x A.

ii) Find the domain and range of R.

36} 25, 16, 9, 4,{1, Range

6} 5, 4,3, 2, {1, Domain

36)} (1, 25) (1, 16) 9)(1, 1)(1,4)(1, {(1, A x A

36} 25, 16, 9, 4,{1, A of Squares

6} 5, 4,3, 2, {1, A

45} 4,...,3, 2, {1, A

=

=

=

=

=

=

6. A Relation R is given by set {(x, y) /y = x + 3,

x ∈ {0, 1, 2, 3, 4, 5}}. Determine its domain

and range.

8 3 5 y5, x

7 3 4 y 4, x

6 3 3 y3, x

5 3 2 y2, x

43 1 y1, x

3 3 0 y0, x

3 x y

=+==

=+==

=+==

=+==

=+==

=+==

+=

8} 7, 6, 5, 4,{3, R of Range

5} 4,3, 2, 1, {0, R ofDomain

8)} (5, 7) (4, 6) (3, 5) (2, 4)(1, 3) {(0, R

=

=

=

7. The arrow diagram shows a relationship

between the sets P and Q. Write the relation

in (i) Set builder form (ii) Roster form (iii)

What is the domain and range of R.

(i) Set builder form

R= {(x,y) |y= x – 2 x ∈ P, y ∈Q}

(ii) Roster form R R = {(5,3), (6,4), (7,5)}

(iii) Domain of R = {5,6,7}

Range of R = {3,4,5}

8. Represent each of the given relations by

(a) an arrow diagram, (b) a graph and (c) a set in

roster form, wherever possible.

(i) {(x, y)| x = 2y, x ∈ {2, 3, 4, 5}, y ∈ {1, 2, 3, 4}

5/2 y5 x

2 4/2 y 4 x

3/2 y3 x

1 2/2 y 2, x

x/2 y2y x

==

===

==

===

=⇒=

Required relation = {(2, 1) (3, 3/2) (4, 2) (5, 5/2)}

(i) Arrow diagram

(ii) A graph

(iii) Roster form R = {(2, 1) (4, 2)}

(ii) {(x, y)|y = x + 3, x, y natural numbers < 10}

9 y6 x

8 y5 x

7 y 4 x

6 y3 x

5 y2 x

4 y1 x

3 x y

==

==

==

==

==

==

+=

R = {(1, 4) (2, 5) (3, 6) (4, 7) (5, 8) (6, 9)}

i) An arrow diagram :

ii) A graph


iii) In Roster form :

R = {(1, 4) (2, 5) (3, 6) (4, 7) (5, 8) (6, 9)}

9. 10 families A, B, C, D, E, F, G, H, I and J with

two children. Among these, families B, F, I

have two girls; D, G, J have one boy and one

girl; the remaining have two boys. Define a

relation R by xRy, where x denote number of

boys and y denote family with x number of

boys. Represent this situation as a relation

through ordered pairs and arrow diagram.

Domain of R = {0,1,2},

where 0, 1, 2 represent no boy, one boy, two boys

Families with two girls are the ones with no boys

R= {(0,B)(0,F)(0,I)(1,D)(1,G)(1,J)(2,A)(2,C)(2,E)(2,H)

10. A company has four categories of employees

given by Assistants (A), Clerks (C), Managers

(M) and an Executive Officer (E). The company

provide Rs.10,000, Rs.25,000, Rs.50,000 and

Rs.1,00,000 as salaries to the people who work in

the categories A, C, M and E. If A1, A2, A3, A4and

A5 were Assistants; C1, C2, C3, C4 were Clerks;

M1, M2, M3 were managers and E1, E2 were

Executive officers and if the relation R is

defined by xRy, where x is the salary given to

person y, express the relation R through an

ordered pair and an arrow diagram.

"x" be the salary given to person

"y" be the set of employees

{(10000, A1)(10000, A2)(10000, A3)(10000, A4)

(10000, A5) (25000, C1) (25000, C2) (25000, C3)

(25000, C4) (50000, M1)(50000, M2)(50000, M3)

(100000, E1) (100000, E1)}

3. FUNCTIONS

1. X={1,2,3,4},Y= {2,4,6,8,10}, R = {(1,2),(2,4),(3,6),

(4,8)} Show that R is a function and find its

domain, co-domain and range?

i) for each x ∈ X , there exists only one y ∈ Y.

All elements in X have only one image in Y.

Therefore R is a function.

ii) Domain X = {1,2,3,4}

Co-domain Y = {2,3,6,8,10}

Range of f = {2,4,6,8}.

2. A relation ‘f’ is defined by f (x) =x2 – 2

where, x ∈ {-2,-1,0,3} (i) List the elements of f

(ii) Is f a function?

}0,-2)(3,7)2)(-1,-1)(,{(

72)()(

22)()(

-2)()(

22)()(

2 )(

2

2

2

2

2

2

33

00

111

22

−=

=−=

−=−=

=−−=−

=−−=−

−=

f

f

f

f

f

xxf

ii) Each element in the domain of f has a unique

image. Therefore f is a function.

3. If X={–5,1,3,4} Y = {a,b,c}, which of following

relations are functions from X to Y ?

(i) R1 = {(–5,a), (1,a), (3,b)}

(ii) R 2 = {(–5,b), (1,b), (3,a),(4,c)}

(iii) R3 = {(–5,a), (1,a), (3,b),(4,c),(1,b)}

i) R1 is not a function as 4∈ X does not have an

image in Y.

(ii) R2 is a function as each element of X has an

unique image in Y.


iii) 1∈X has two images a ( Y and b ( Y.

Image of an element should always be unique

R3 is not a function

4. Let f = {(x, y) | x, y ∈ N and y = 2x} be a

relation on N. Find the domain, co-domain

and range. Is this relation a function?

Since x and y ∈ N,

8 2(4) y 4 x

6 2(3) y3 x

4 2(2) y2 x

2 2(1) y1 x

===

===

===

===

f = {(1, 2) (2, 4) (3, 6) (4, 8)................}

For each values of x, we get different values of y.

So the given relation is a function.

.....} 8, 6, 4,{2, Range

...}.......... 4,3, 2, {1, domain Co

..}.......... 4,3, 2, {1, Domain

=

=

=

5. Let X = {3, 4, 6, 8}. Determine whether the

relation R = {(x, f (x)) | x ∈ X, f (x) = x2 +

1} is a function from X to N ?

f (x) = x2 + 1

65 18 f(8)8 x if

37 16 f(6)6 x if

17 1 4 f(4) 4 x if

10 13 f(3)3 x if

2

2

2

2

=+==

=+==

=+==

=+==

R = { (3, 10) (4, 17) (6, 37) (8, 65) }

For each values of x, we get different values of

f(x). Hence it is a function

6. Given 22 )( xxxf −= ,

find (i)f (1) (ii) f (x+1) (iii) f (x) + f (1)

121

1

1222

112

11

2

2

2

+−=+

+−=

−−−+=

+−+=+

=−=

−=

2

2

2

)()(

)()( 1)f(x

)(1)(2 (1) f

2 )(

xxfxf

x

xxx

xx

xxxf

7. Given function f : x → x2 −5x + 6 , evaluate

(i) f (-1) (ii) f (2a) (iii) f (2) (iv) f (x −1)

12 7x - x

6 5 5x - 1 2x - x

6 1)-5(x- 1)-(x 1)-f(x (iv)

0

6 10 - 4

6 5(2)- (2) f(2) (iii)

6 10a - 4a

6 5(2a)- (2a) f(2a) (ii)

12

6 5 1

6 5(-1)- (-1) f(-1) (i)

6 5x- x f(x)

2

2

2

2

2

2

2

2

+=

+++=

+=

=

+=

+=

+=

+=

=

++=

+=

+=

8. A graph representing function f (x) is given

it is clear that f (9) = 2.

(i) Find the following values of the function

(a) f (0) (b) f (7) (c) f (2) (d) f (10)

(ii) For what value of x is f (x) = 1?

(iii) Describe following (i) Domain (ii) Range.

(iv) What is the image of 6 under f ?

(i) (a) f(0) = 9, (b) f(7) = 6,

(c) f (2) = 6, (d) f (10) = 0

(ii) For x = 9.5, we get 1.

(iii) Domain = { 0 ≤ x ≤ 10 }

Range = { 0 ≤ x ≤ 9 }

(iv)


9. Let f (x) = 2x + 5. If x ≠ 0 find x

f(2) - 2) f(x +

2

x

9 - 9 2x

x

f(2) - 2) f(x

9

5 2(2) f(2)

2x

5 2) 2(x 2) f(x

5 2x (x) f

=

+=

+

=

+=

+=

++=+

+=

9

10. A function f is defined by f (x) = 2x – 3

(i) find [f(0) + f(1)]/2

(ii) find x such that f (x) = 0.

(iii) find x such that f (x) = x .

(iv) find x such that f (x) = f (1−x) .

21

x

2 4x

1 - 2x- 3 - 2x

3 - x) -2(1 3 - 2x

. x)-(1 f (x) f (iv)

3 x

x 3 - 2x

. x (x) f (iii)

23

x

0 3 - 2x

0. (x) f (ii)

2-

(-1) 3-

2

f(1) f(0)

1- 3 - 2(1) f(1)

3- 3 - 2(0) f(0) (i)

3 2x f(x)

=

=

=

=

=

=

=

=

=

=

=

=

+=

+

==

==

−=

2

11. An open box is to be made from a square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown. Express the volume V of the box as a function of x.

Since the original shape is square, length of all

sides will be equal.

576x 96x - 4x (x) Vcuboid of volume

] 4x 96x - [576 x

](2x) (2x) 2(24) - [24 x

2x) - x(24

2x) - (24 2x) - (24

height width length cuboid of Volume

x height

2x - 24 width length

23

2

22

2

+=

+=

+=

=

=

××=

=

==

x

12. A function f is defined by f (x) = 3−2x . Find

x such that f (x2) = (f (x))2 .

1 x

0 1 - x

0 1) - (x

0 1 2x - x

0 6 12x - 6x

0 3 - 9 12x - 2x 4x

4x 12x - 9 2x - 3

given][(x)) (f )(x f

4x 12x - 9 (x)) (f

(2x) 2(3)(2x) - 3

2x) - (3 (x)) (f

2x - 3 )(x f

2x - 3 (x) f

2

2

2

22

22

22

22

22

22

22

=

=

=

=+

=+

=++

+=

=

+=

+=

=

=

=

13. A plane is flying at a speed of 500 km per

hour. Express the distance d travelled by the

plane as function of time t in hours.

t 500 distance Required

500t d500d

t

hoursin t taken time

d travelled Distance

hour per km 500 plane of Speed

speed / Distance Time

=

=

=

=

=

=

=


14. The data depicts length of a woman’s

forehand and her corresponding height. A

student finds a relationship between height

(y) and forehand length(x) as y = ax +b ,

where a, b are constants.

i) Check if this relation is a function.

ii) Find a and b.

iii) Find the height of a woman whose forehand

length is 40 cm.

iv) Find the length of forehand of a woman if

her height is 53.3 inches.

(i) For every values of x,we get different y values

Hence it is a function.

24.5 0.9x y

24.5 b

40.95- 65.5 b

65.5 b 45.5(0.90)

(1),in 0.90 a Substitute

0.90 a

9.5/10.5 a

9.5 10.5a

9.5 b - b 35a - a 45.5

56 - 65.5 b) (35a - b) a (45.5

(2) - (1)

-(2)- b a 35 56

b a(35) 56

56 y35, x

-(1)- b a 45.5 65.5

b a(45.5) 65.5

65.5 y 45.5, x

b x a y

+=

=

=

=+

=

=

=

=

=+

=++

+=

+=

==

+=

+=

==

+=

(iii) Find height when forehand length is 40 cm.

inches. 60.5 woman of Height

60.5 y

24.5 0.9(40) y

40 x if ? y

=

=

+=

==

(iv) length of forehand if her height 53.3 inches

32 x

.

28.8x

0.9x 24.5 - 53.3

24.5 0.9x 53.3

53.3 yif ? x

=

=

=

+=

==

90

15. Find the domain of given functions

i) 1

1+

=x

xf )( ii) 65

12 +−

=xx

xf )(

} {2,3 ofdomain So

2,3 x except numbers real all for defined is

defined. not are (3) f and (2) fthen ,2,3 x If

)(ii)

}{R ofdomain

1- x except numbers real all for defined is

defined not is (-1) fthen 1- x If

)(

)(i)

−=

=

=+−

=

−−=

=

=

=−

+=

Rf

f

xxxf

f

f

f

xxf

65

1

1

01

1

11

2

4. TYPES OF FUNCTIONS

1. Let A = {1,2,3,4} and B = {2,5,8,11,14} be two sets.

Let f: A →B be a function given by f (x) = 3x−1.

Represent this function (i) arrow diagram (ii)

table form (iii) set of ordered pairs (iv) in a

graphical form

(i) Arrow diagram

111434

81333

51232

21131

13

=−=

=−=

=−=

=−=

−=

)()(

)()(

)()(

)()(

)(

f

f

f

f

x xf

(ii) Table form

x 1 2 3 4

f(x) 2 5 8 11

(iii) Set of ordered pairs :

f = {(1,2),(2,5),(3,8),(4,11)}

(iv)Graphical form


2. Let f : A → B be a function defined by

1 - 2x

)( =xf where A = {2, 4, 6, 10, 12}, B = {0, 1, 2,

4, 5, 9} . Represent f by

(i) set of ordered pairs; (ii) table;

(iii) arrow diagram; (iv) a graph

51 - 2

12 f(12)

4 1-2

10 f(10)

2 1 - 26

f(6)

11 - 24

f(4)

01 - 22

(2)

1 - 2x

)(

==

==

==

==

==

=

f

xf

i) Set of ordered pairs = {(2,

0)(4,1)(6,2)(10,4)(12,5)}

ii) a table

iii) an arrow diagram;

iv) a graph

3. Represent the function f = {(1, 2),(2, 2),(3, 2),(4,

3),(5, 4)} through

(i) arrow diagram (ii) a table form (iii) a graph

(i) an arrow diagram

(ii) a table form

(iii) a graph

4. Determine whether graphs given below

represent functions. Give reason

i) Curve does not represent a function as the

vertical line meets curve in two points P and Q

ii) Curve represents a function as the vertical lines

meet the curve in at most one point

iii) Curve does not represent a function as vertical

lines meet the curves in two points P and Q

iv) Curve represents a function as the vertical lines

meet the curve in at most one point


v) Since the graph intersects the vertical line (y-axis)

at two points, it is not a function.

vi) given graph intersects the vertical line (y-axis) at

one point. It is a function.

vii) Graph intersects the y-axis at three points,

hence it is not a function.

viii) The graph intersects the vertical line at most

one point. Hence it is a function.

5. Determine type of function.Justify your Answer

i) One-One and Onto (bijection)

Distinct elements of A have distinct images in B

and every element in B has a pre-image in A.

ii) One to One

Distinct elements of A have distinct images in B

iii) Many to One

Two or more elements of A have same image in B

iv) Onto

Range of f = co-domain

Every element in B has a pre-image in A

v) Into

Range of f is a proper subset of co-domain

There exists at least one element in B which is

not image of any element of A

6. Using horizontal line test determine which of

the following functions are one – one.

i) Curve represents a one–one function as hori

zontal line meet curve in only one point P


ii) Curve does not represent one–one function, as

horizontal line meets curve in two points P ,Q.

iii) Curve represents one–one function as hori

zontal line meets curves in only one point P.

7. Check whether a function. If so mention the

type. Justify your answer

i) A= {1,2,3,4}, B= {a,b,c,d,e}

f = {(1,a), (2,b), (3,d), (4,c)}

ii) A= {1,2,3,4}, B= {a,b,c,d,e}

g = {(1,a), (2,b), (3,c), (4,e)}

iii) A = {1,2,3,4}, B= {a, b, c}

f = {(1,a), (2,a), (3,b), (4,c)}

iv) A= {x,y,z}, B= {l,m,n}

f= {(x,m)(y,n)(z,l)}

v) A = {1,2,3,} , B= {w,x,y,z},

f = {(1,w) (2,z) (3,x)}

i) For different elements in A, there are different

images in B.

Hence f is a one – one function.

ii) g is a function from A to B.

Two distinct elements 1 and 2 in first set A

have same image b in the second set in B.

Hence, g is not a one–one function.

iii) f is a function from A to B

Different elements 1 and 2 of A have same

image a in B.

Hence f is a many – one function.

iv) Range of f ={l,m,n} = B

Hence f is an onto function.

v) y ∈ B is not image of any element in A.

Here, range of f = {w,x,z} ⊂ B

Therefore, f is a into function.

8. Let A = {1,2,3} , B = {4,5,6,7} and f = {(1,4)} ,( 2,5),( 3,6)}

be a function from A to B. Show that f is one –

one but not onto function.

For different elements in A, there are different

images in B.

∴ f is one–one function.

Element 7 in the co-domain does not have any

pre-image in the domain.

∴ f is not onto

Therefore f is one–one but not an onto

function.

9. If A = {-2,-1,0,1,2} f: A → B is an onto function

defined by 12 ++= x xf (x) then find B.

71222

31111

11000

11111

31222

1

2

2

2

2

2

2

=++=

=++=

=++=

=+−+−=−

=+−+−=−

++=

)()()(

)()()(

)()()(

)()()(

)()()(

f

f

f

f

f

x xf (x)

Since, f is an onto function,


Range of = co-domain of f.

Therefore, B= {1,3,7}

10. Show that the function f : N → N defined by f

(x) = 2x – 1 is one-one but not onto

If for all a1, a2 ∈ A, f(a1) = f(a2) implies a1 = a2 then

f is one – one function.

y x

2y 2x

1 - 2y 1 - 2x

f(y) f(x)

N, yx, Let

=

=

=

=

∈

Hence the function is one to one.

It is not onto :

co-domain of the function = range of function

Even numbers in co-domain are not associated

with the elements of domain.

Hence it is not onto.

11. Show that the function f : N → N defined by f

(m) = m2 + m + 3 is one-one function.

y x

0 y- x

0 1) y (x y)- (x

0 y)- (x y)- (x y) (x

0 y- x y- x

y y x x

3 y y 3 x x

f(y) f(x)

3 m m (m) f

f(y) f(x) N, yx, Let

22

22

22

2

=

=

=++

=++

=+

+=+

++=++

=

++=

=∈

Hence it is one to one function.

12. Let f be a function Ν→Ν:f , defined by

,)( Ν∈+= xxxf 23

i) Find the images of 1, 2, 3

ii) Find the pre-images of 29, 53

ii) Identify the type of function

i) ,)( Ν∈+= xxxf 23

112333

72232

52131

=+=

=+=

=+=

)()(

)()(

)()(

f

f

f

Images of 1, 2, 3 are 5, 8, 11

(ii) If x is the pre-image of 29 or 53

9

273

2923

29

=

=

=+

=

x

x

x

xf )(

17

213

5323

53

=

=

=+

=

x

x

x

xf )(

Pre-images of 29 and 53 are 9 and 17

iii) Since different elements of have different

images in the co-domain, the function f is one

– one function.

But range of f = {5, 8, 11, 14, 17, ...} is a proper

subset of .

Therefore f is not an onto function.

That is, f is an into function.

Thus f is one – one and into function.

13. Let A = {1, 2, 3, 4} and B = N . Let f : A → B be

defined by f (x) = x3 then, (i) find the range of f

(ii) identify the type of function

64} 27, 8, {1, f of Range

64 4 (4) f 4 x

27 3 (3) f3 x

8 2 (2) f2 x

1 1 (1) f1, x

x (x) f

3

3

3

3

3

=

===

===

===

===

=

Every element in A has associated with different

elements of B. Hence it is one to one.

14. In each of the following cases state whether the

function is bijective or not. Justify your answer.

i) f : R → R defined by f (x) = 2x +1

ii) f : R → R defined by f (x) = 3 – 4x2

i) Testing whether it is one to one :

y x

2y 2x

1 2y 1 2x

f(y) f(x)

f(y) f(x) R, yx, Let

=

=

+=+

=

=∈

So, it is one to one.

Testing whether it is onto :

If f :A→ B is onto function then, range of f = B

y f(x)

1 1 - y f(x) 2

1) -2(y f(x)

2

1 - y x

1 2x y

1 2x f(x) y

B (A) f i.e

=

+=

+=

=

+=

+==

=

1


It is onto function. Hence it is bijective function.

(ii) f : R → R defined by f (x) = 3 – 4x2

Testing whether it is one to one : Let x, y ∈ R,

y- x (or) y x

0 y x (or) 0 y- x

0 y) y)(x- (x

0 y- x

4y 4x-

4y3 4x-3

f(y) f(x)

22

22

22

==

=+=

=+

=

=

−=

=

It is not one to one.

Hence it is not bijective function.

15. Let A = {−1, 1}and B = {0, 2} . If the function f : A

→ B defined by f(x) = ax + b is an onto function?

Find a and b.

1 a

1- a-

0 1 a- (1),in b Sub

1 b

2 2b(2) (1)

-(2)----- 2 b a

2 b a(1)

2 f(1)

-(1)----- 0 b a-

0 b a(-1)

0 f(-1)

b ax f(x)

=

=

=+

=

=+

=+

=+

=

=+

=+

=

+=

16. If the function f : R→ R is defined by

≥−

≤≤−−

−<+

=

323

322

2722

xifx

xifx

xifx

xf )( Find

values of (i) f (4) (ii) f (-2) (iii) f (4)+2 f(1) iv)

)(

)()(

3

431

−

−

f

ff

2222

2

2

102434

23

4

2

2

=−−=−

−=

−=

=−=

−=

=

)()(

)(

interval. second thein lie ii)

)()(

)(

interval. third thein lie i)

f

xxf

x

f

xxf

x

311

1031

3

431

17323

72

3

81210124

1211

2

1

2

2

−=−−

=−

−

=+−=−

+=

−=

=−+=+

−=−=

−=

=

)(

)(

)()(

)()(

)(

interval. first thein lie iv)

)()()(

)()(

)(

interval. second thein lie iii)

f

ff

f

xxf

x

ff

f

xxf

x

17. If the function f is defined by

−<<−−

≤≤−

>+

=

131

112

12

xifx

xif

xifx

xf )(

Find the values of

(i) f (3) (ii) f (0) (iii) f (−1.5) (iv) f (2)+ f (−2)

(i) f(3)

5 f(3)

2 3 f(3)

2 x f(x)

=

+=

+=

(ii) f(0) 0 lies between -1 and 1.

f (0) = 2.

(iii) f (−1.5)

2.5- f(-1.5)

1 - 1.5- f(-1.5)

1 - x f(x)

=

=

=

(iv) f (2)+ f (−2)

1

(-3) 4 (-2) f (2) f

3-

1 - 2- f(-2)

1 - x f(x) Thus,

1xin lies 2-

4

2 2 f(2)

2 x f(x)Thus,

1xin lies 2

=

+=+

=

=

=

<

=

+=

+=

>

18. A function f : [−5,9] → R is defined as follows:


<<−

<≤−

<≤−+

=

9643

6215

25162

xifx

xifx

xifx

xf )( Find

(i) f (−3) + f (2) (ii) f (7) - f (1) (iii) 2f (4) + f (8)

(iv) f(-2) f(4)

f(6) -2f(-2)

+

(i) f (−3) + f(2)

219 17 - f(2) (-3) f

191 - 5(2) f(2)

f(2) for 1 - 5x f(x)

17- 1 6(-3) f(-3)

f(-3) for 1 6x f(x)

2

2

=+=+

==

=

=+=

+=

(ii) f (7) - f (1)

10 7 -17 (1) f - (7) f

1 6(1) f(1)

f(1) for 1 6x f(x)

17 4- 3(7) f(7)

f(7) for 4- 3x f(x)

==

=+=

+=

==

=

7

(iii) 2f (4) + f (8)

178

20 2(79) (8) f (4) 2f

20 4- 3(8)

4- 3x f(8)

f(8) for 4- 3x f(x)

79 1 - 80

1 - 5(4) f(4)

f(4) for 1 - 5x f(x)2

2

=

+=+

==

=

=

==

=

=

(iv) f(-2) f(4)

f(6) - 2f(-2)

+

179-

(-11) 79

14 - 2(-11)

f(-2) f(4)

f(6) - 2f(-2)

14 4- 3(6) f(6)

f(6) for 4- 3x f(x)

791 - 5(16) f(4)

f(4) for 1 - 5x f(x)

11- 1 6(-2) f(-2)

f(-2) for 1 6x f(x)

2

2

=

+=

+

==

=

==

=

=+=

+=

19. Forensic scientists can determine height of a

person based on the length of their thigh bone.

They do so using function bh (b) 1054472 .. +=

where b is the length of the thigh bone.

(i) Check if the function h is one – one

(ii) Find height of a person if the length of his

thigh bone is 50 cms.

(iii) Find the length of the thigh bone if the

height of a person is 147.96 cms.

21

21

21

21

472472

10544721054472

1054472

bb

bb

bb

bhbh

bh (b)

=

=

+=+

=

+=

..

....

)()( Assume,

..

So, the function h is one – one.

ii) If the length of the thigh bone b = 50

6177

10545047250

.h Height

.)(.

=

+= )h (

iii) If the height of a person is 147.96 cms,

cmsLength ..

...

.)(

384728693

961471054472

96147

=

=

=+

=

b

b

b

bh

20. The distance S (in kms) travelled by a particle in

time ‘t’ hours is given by 2

2 ttts

+=)( . Find the

distance travelled by the particle after

(i) three and half hours.

(ii) eight hours and fifteen minutes.

(i) three and half hours i.e t = 3.5 hrs

kms.

.).().(

87572

535353

2

=

+=s

(ii) 8 hours and 15 minutes i.e t = 8.25 hrs

kms.

.).().(

15625382

258258258

2

=

+=s

21. Distance S an object travels under the influence

of gravity in time t seconds is given by S(t) =

(1/2) gt2 + at + b

where, g is acceleration due to gravity, a, b are

constants. Check if the function S(t) is one-one.

If for all a1, a2 ∈ A, f(a1) = f(a2) implies a1 = a2 then

f is called one – one function.


yx

x - y

yx x - y

x - y

baygybax gx

b ay gy baxgx

=

=

=++

=++

=++−++

++=++

=

=∈

0

0 a] )g( [ )(

0 b - b )a( ) y- g(x

S(y) S(x)

f(y) yx, Let

22

21

21

021

21

21

21

22

22

Hence it is one to one function.

22. Function ‘t’ which maps temperature in Celsius

(C) into temperature in Fahrenheit (F) is defined

by t(C) = F where 32 C 59

F += . Find,

(i) t(0) (ii) t(28) (iii) t(-10)

(iv) value of C when t (C) = 212

(v) temperature when Celsius value is equal to

Farenheit value.

C 100 C of value

32 - 212 C

212 32 C

212 (C) t (iv)

F 14

3218-

32 (-10) t(-10) (iii)

F 82.4

32 (28) t(28) (ii)

F 32

32 (0) t(0) (i)

)(

°=

=

=+

=

°=

+=

+=

°=

+=

°=

+=

+=

59

59

59

59

59

3259

C Ct

(v) temperature when Celsius =Farenheit value.

F = C

Hence answer is -40°

23. Identify the type of function

i)A= { a,b,c,d }, B = {1,2,3} , f = {(a,3)(b,3)(c,3)(d,3)}

ii) A = { a, b, c }, f = {(a, a) (b, b)( c, c)}

i) Since, f(x) = 3 for every x ∈ A , Range of f =

{3}

f is a constant function.

ii) Since f (x) = x for all x ∈ A

It is identity function on A.

24. f is a function from R to R defined by f (x) = 3x – 5

Find values of a and b given that (a, 4) and (1, b)

belong to f.

f (x) = 3x – 5 written as f = {( x, 3x – 5)|x ∈ R}

(a, 4) means image of a is 4.

3 a

4 5-3a

4 (a) f

=

=

=

(1, b) means the image of 1 is b.

2 b

b 5 -3(1)

b (1) f

−=

=

=

5. COMPOSITION OF FUNCTIONS

1. Find fog and gof if f (x) = 2x+1, g (x) = x2 – 2

144

212

12

32

122

2

2

2

2

2

2

−+=

−+=

+=

−=

+−=

−=

xx

x

xggof

x

x

xffog

)(

)(

)(

)(

2. If f (x) = 3x−2 , g(x) = 2x+k and if fo g =go f then find the value of k.

1

46236

232223

−=

+−=−+

+−=−+

=+

=

+==

k

kxkx

kxkx

g

)()(

2)-3x(k)f(2x

f gog fo

k 2x g(x) ,2-3x (x) f

3. f (x) = 2x+3 g (x) = 1 −2x, h (x) = 3x¸ Prove that (fo go h) = fo( goh)


x

x

xfgohfo

x

x

xggoh

x

x

xfoghfog

x

x

xffog

125

3612

61

61

321

3

1125

345

3

45

3212

21

−=

+−=

−=

−=

−=

=

−=

−=

=

−=

+−−=

−=

)(

)()(

)(

)(

).....(

)(

)()(

)(

)(

4. Find x if gff(x) = fgg(x), given f x (x) = (3x +1) and gx (x) = x + 3

.

2x

126x

19 + 3x 7 + 9x

] 1 + 6) + 3(x [ 3] + 4)+ (9x [

6) + (x f 4)+ (9xg

3] + 3) + (x [ f 1]+1)+ 3(3x [g

3)] + (xg [ f 1)]+ (3x f [g

(x)}]{g [g f (x)}] {f [fg

fgg(x)= gff(x)

=

=

=

=

=

=

=

=

5. Using the functions f and g, find f o g and g o f . Check whether f o g = g o f . (i) f (x) = x −6, g(x) = x2

(x) f og (x)g o f

6) - (x

g[f(x)] (x) f og

6 - x

f[g(x)] (x)g o f

2

2

≠

=

=

=

=

(ii) f (x) = 2/x, g(x) = 2x2 - 1

(x) f og (x)g o f

1-x

8

1 - 2

2

g[f(x)] (x) f og

1 - 2x

2

1] - f[2x (x)g o f

2

2

2

2

≠

=

=

=

=

=

x

iii) f (x) = (x + 6)/3, g(x) = 3 - x

3

x-9

3

6x-3

x] - f[3 (x)g o f

=

+=

=

(x) f og (x)g o f3

x-3

3

6 x - 3

3

6 xg (x) f og

≠

=

+=

+=

(iv) f (x) = 3 + x, g(x) = x - 4

(x) f og (x)g o f

(2)--- 1 - x

4- x 3

x] [3g (x) f og

1 - x

4- x 3

4]- f[x (x)g o f

=

=

+=

+=

=

+=

=

(v) f (x) = 4x2 − 1, g(x) = 1 + x

(x) f og (x)g o f

4x

1 - 4x 1

1] - [4xg (x) f og

3 8x 4x

1 - 2x) x 4(1

1 - x) 4(1

x] f[1 (x)g o f

2

2

2

2

2

2

≠

=

+=

=

++=

++=

+=

+=

6. Find the value of k, such that f o g = g o f

(i) f (x) = 3x +2, g(x) = 6x −k

5- k

10 2k -

12 2- k 3k -

k - 12 18x 2 3k - 18x

k - 2) 6(3x 2 k) - 3(6x

2] g[3x k] - f[6x

g[f(x)] f[g(x)]

f og g o f

=

=

+=+

+=+

+=+

+=

=

=

(ii) f (x) = 2x −k, g(x) = 4x + 5

35

- k

5- 3k

10 - 5 4k k -

5 4k - 8x k - 10 8x

5 k) - 4(2xk - 5) 2(4x

k] - g[2x 5] f[4x

g[f(x)] f[g(x)]

f og g o f

=

=

=+

+=+

+=+

=+

=

=


7. f (x) = 2x −1,g (x) = 2

1 x +, show that fog=gof = x

-(2)- x 2

1) 1 - (2x

1] - g[2x f og

-(1)- x

1 - 2

1 x2

f g o f

=

+=

=

=

+

=

+

=2

1 x

(i) If f (x) = x2 −1, g(x) = x −2 find a, if g o f (a) = 1

2 a

4 a

1 3- a

1 (a) f og

: thatGiven

3- a (a) f og

2 - 1)- (a

1]- g[a (a) f og

2- a g(a)

1- a f(a)

2

2

2

2

2

2

±=

=

=

=

=

=

=

=

=

(ii) Find k, if f (k) = 2k −1 and f o f (k) = 5.

. 2 k

8 4k

3 5 4k

5 3 -4k

5 (k) f o f

: thatGiven

3 - 4k

1 - 1) -2(2k

1] -f[2k (k) f o f

=

=

+=

=

=

=

=

=

8. Let A, B, C ∈N and function f:A → B be defined

by f(x) = 2x + 1 and g:B → C be defined by g(x) = x2 . Find range of f o g and g o f.

N} x and 1) (2x y| {y

: Range

1) (2x

1] g[2x f og

N} x and 1 2x y| {y

: Range

1 2x y

1 x 2

]f[x g o f

2

2

2

2

2

2

∈+=

+=

+=

∈+=

+=

+=

=

9. Let f (x) = x2 −1 . Find (i) f o f (ii) f o f o f

1 - )2x - (x

]2x - f[x

f] o [f f f o f o f (ii)

2x - x f o f

1 - 1 2x - x

1 - 1)- (x

1]- f[x f o f (i)

224

24

24

24

22

2

=

=

=

=

+=

=

=

10. If f : R → R and g : R → R are defined by f(x) = x5

and g(x) = x4 then check if f, g are one-one and f

o g is one-one?

f(x) = x5

For every positive and negative values of x, we get positive and negative values of y.

Every element in x is associated with different elements of y. Hence it is one to one function.

g(x) = x4

For every positive and negative values of x, we get only positive values of y.

Negative values of y is not associated with any elements of x. Hence it is not one to one function.

fog(x) = f[g(x)]

= f[x4]

now, we apply x4 instead of x in f(x)

f[x4] = (x5)4

fog(x) = x20

fog is not one to one function.

11. Consider functions f (x), g(x), h(x) given below.

Show that (f o g) o h = f o (g o h) in each case.

(i) f(x) = x −1, g(x) = 3x +1 and h(x) = x2

h) o(g o f h o g) o (f

(2) (1)

(2)------- 3x h) o(g o f

1 - 1 3x

1] f[3x h) o(g o f

1 3x

]g[x h) o(g

:h) o(g o f

-(1)------- 3x

][x g) o (f h o g) o (f

3x

1 - 1 3x

1] f[3x fog(x)

:h o g) o (f

2

2

2

2

2

2

2

=

=

=

+=

+=

+=

=

=

=

=

+=

+=


(ii) f (x) = x2, g(x) = 2x and h(x) = x + 4

(2)-------64 x 32 4x

8 2(2x)(8) (2x)

8) (2x

8] f[2x h) o(g o f

8 2x 4) 2(x

4] g[x h) o(g

-(1)------- 64 32x 4x

16) 8x 4(x

4) 4(x

4] [x g) o (f h o g) o (f

4x (2x)

f[2x] fog(x)

2

22

2

2

2

2

22

++=

++=

+=

+=

+=+=

+=

++=

++=

+=

+=

==

=

(iii) f (x) = x −4, g(x) = x2 and h(x) = 3x −5

(2) (1)

-(2)------- 21 30x - 9x

25] 30x - f[9x h) o(g o f

25 30x - 9x

5)- (3x

5]- g[3x h) o(g

-(1)------- 21 30x - 9x

4- 5 (3x)(5) 2 - (3x)

4- 5) - (3x

5]- [3x g) o (f h o g) o (f

4- x

]f[x fog(x)

2

2

2

2

2

22

2

2

2

=

+=

+=

+=

=

=

+=

+=

=

=

=

=

12. Let f = {(−1, 3),(0,−1),(2,−9)} be a linear function from Z into Z . Find f (x).

1. - 4x- equation Rqd

and of valuetheApplying

4- a

4 a-

1 3 a-

3 (-1) a-

(1)in of valueApplying

1- b

b 1-

b a(0) 1-

1- then y 0, x if

(1)------- 3 b a-

b a- 3

b a(-1) 3

3 then y 1,- x if

function Linear

=

=

=

+=

=+

=

=

+=

==

=+

+=

+=

==

+=

y

ba

b

b ax y

13. In electrical circuit theory, a circuit C(t) is called

a linear circuit if it satisfies the superposition

principle given by C(at1 + bt2) = aC(t1) + bC(t2),

where a,b are constants. Show that the circuit

C(t) = 3t is linear.

Take two points t1 and t2 from domain of C(t).

c(at1 + bt2) = 3(at1 + bt2)

c(at1) = 3at1

ac(t1) = 3at1

bc(t2) = 3bt2

3(at1 + bt2) = 3at1 + 3at2

(or) C(at1 + bt2) = aC(t1) + bC(t2),

Hence c(t) is linear.

5. EUCLID’S DIVISION LEMMA

1. Find all positive integers, when divided by 3

leaves remainder 2.

............ 14, 11, 8, 5, 2,

8 = a

2 + 3(2) = a

(given) 3 = b ,2 = r 2, = q let

5 = a

2 + 3(1) = a

(given) 3 = b ,2 = r 1, = q let

2 = a

2 + 3(0) = a

(given) 3 = b ,2 = r 0, = q let

r + bq = a algorithm,division By

2. A man has 532 flower pots. He wants to arrange

them in rows such that each row contains 21

flower pots. Find the number of completed rows

and how many flower pots are left over.

Total number of flower pots = 532

Number of flower pots in each row = 21

We may arrange the flower pots in 25 rows with

each row consists of 21 pots.

The remaining number of flower pots = 5.

3. Prove that the product of two consecutive

positive integers is divisible by 2.

Let "x" be the positive integer

consecutive number be "x + 1"

When a positive integer is divided by 2 the

remainder is either 0 or 1. So, any positive integer

will of the form 2k, 2k+1 for some integer k.


1) +2k(2k 1) + (x x

2k =x

numbereven = x

: 1 Case

=

Hence it is divisible by 2.

1) +1)(2k +(k 2 =

2) +1)(2k +(2k =

1) + 1 +1)(2k +(2k = 1) + (x x

1 +2k = x

number odd = x If

: 2 Case

So, the product is divisible by 2.

4. When the positive integers a, b and c are divided by 13, the respective remainders are 9,7 and 10. Show that a+b+c is divisible by 13.

13 by divisible is c + b + a Hence

2] + )q + q + 13[(q =

13(2) + )q + q + 13(q =

26 +13q + 13q + 13q =

10 + 13q +7 + 13q + 9 + 13q = c + b + a

(3) + (2) + (1)

-(3)---10 + q 13 = c

-(2)---7 + q 13 = b

-(1)--- 9 + q 13 = a

remainder + quotient x divisor = divided

321

321

321

321

3

2

1

.

5. Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.

1. remainder leaves and 4by divisible isn

1 + m) + 4(m= n

1 + 4m+ 4m=

12 + 2(2m) + (2m) =

1) + (2m = n

1+2m =n let

odd... isn : 2 Case

0. remainder leaves and 4by divisible isn

4m= n

2m =n let

integereven isn : 1 Case

22

2

2

22

22

∴

∴

6. Use Euclid’s Division Algorithm to find the Highest Common Factor (HCF) of i) a = 273 and b = 119 .

7 55) ,(210 of HCF

0 +2 ×7 = 14

7+2× 14 = 35

114+3 ×35 = 119

35+ 2× 119 = 273

=

ii) a = 340 and b = 412

0 + 2(4) = 8

4+ 1(8) = 12

8 + 1(12) = 20

12 + 2(20) = 52

20 + 1(52) = 72

52 + 4(72)= 340

72 + (340) 1 = 412

340 > 412

Hence the required H.C.F is 4. iii) 867 and 255

0 + 2(51) = 102

51 + 2(102) = 255

102 + 3(255) =867

255 >867

255 = b and867 = a

Hence the H.C.F is 51.

iv) 10224 and 9648

0 + 3(144) = 432

144 + 1(432) = 576

432+ 16(576) = 9648

576 + 1(9648) = 10224

9648 > 10224

Hence the H.C.F is 144.

v) 84, 90 and 120

0 + 6(14) = 84

6 + 1(84) = 90

84 > 90

Hence the required H.C.F is 6. 7. Find the largest number which divides 1230 and

1926 leaving remainder 12 in each case.

0 + 174(3) = 522

174 + 522(1) = 696

522 + 696(1) = 1218

696 + 1218(1) = 1914

1914. and 1218 of H.C.F : find To

1914 = 12 - 1926

1218 = 12 - 1230

Rqd number is 174.

8. If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y .

4.is 32 and 60 of H.C.F

-(3)--- 0 + 4(7)= 28

-(2)--- 4+ 28(1) = 32

-(1)--- 28 + 32(1) = 60

32 > 60

∴


1.- = yand 2 = x Hence

(-1) 60 + 32(2) = 4

1 ? 60 - 1) + 32(1 = 4

1 ? 60 - (1) 32 + 32 = 4

1 ? (1) 32 + 1 ? 60 - 32 = 4

1 ? 32(1)) - (60 - 32 = 4

(4)in (5) from 28 of valueApply

(5)---32(1) - 60 = 28 (1), From

(4)--- 28(1) - 32 = 4(2), From

y60 + x 32 = 4

60y + 32x = d

4= d

9. A positive integer when divided by 88 gives the

remainder 61. What will be the remainder when

the same number is divided by 11?

11 by dividedwhen 6 remainder gives

11 by divisible is 55

11. by divisible is 6)-55)-((n

11. by divisible is 61)-(n

88. by divisible is 61)-(n

number. integer that be n"" Let

n

∴

∴

∴

10. Prove that two consecutive positive integers are always coprime.

prime.-co are 1 + x & x :ioncontradict by So

1 =n or 1 dividesn

x - 1 + x dividesn

1. + x and x dividesn

n, = 1) + x (x, of H.C.F

1 >n assume

1 + x x, : numbers econsecutiv

∴

∴

11. Find quotient,remainder when a is divided byb [Find q and r for a and b satisfying a=bq+r]

(i)a =−12 , b = 5

3

3

33512

=

−=

+−=−

+=

r

q

rbqa

Remainder

Quotient

)(

:lemmadivision Euclid By

(ii)a = 17 , b =−3

2

5

25317

=

−=

+−−=

+=

r

q

rbqa

Remainder

Quotient

))((


(iii)a =−19 , b =−4

1

5

15419

=

=

+−=−

+=

r

q

rbqa

Remainder

Quotient

))((


12. Show that the square of an odd integer is of the form 4q +1 , for some integer q.

)( where,

)(

1) +2k(x

k. integer some for 1, +2k= x

integer,even than more one integer odd

integer. odd any x

22

1

14

114

144 2

+=

+=

++=

++=

=

=

=

kkq

q

kk

kk

13. If the Highest Common Factor of 210 and 55 is expressible in the form 55 x -325 , find x.

6 x

5325-55x

5 55) ,(210 of HCF

0 +2 × 5 = 10

5+4× 10 = 45

10+1 × 45= 55

45+ 3 × 55 = 210

=

=

=

14. Find the greatest number that will divide 445 and 572 leaving remainders 4 and 5 respectively.

63. is number required

63 = 441,567 of HCF

02× 63= 126

633 ×126 = 441

1261× 441=567

567. and 441of HCF Find

7 556- ,5724414- 454

∴

∴

+

+

+

∴

==

15. Find the HCF of 396, 504, 636.

12. 636 and 504 396, HCF

12 = 636,36 of HCF Thus

02× 12 = 24

121× 24 = 36

2417×36 = 636

:36 ,636 of HCF find To

= 36,396,504 of HCF hus

02× 36 = 72

361× 72 = 108

723 ×108 = 396

1081×396 = 504

504 and 396: of HCF find To

=

+

+

+

=

+

+

+

+

T


6. FUNDAMENTAL THEOREM OF ARITHMETIC

1. In the given factor tree, find numbers m and n.

50 =n 300, = m are numbers Rqd

300=150×2 = m of Value

50=501×3 = bottom from box 2

50=10×5 =n of Value

10= 2×5 = bottom from box 1

nd

st

2. Is 7x5 x3x2 +3 a composite number? Justify your

answer.

number composite a is it

primes two toin factorized is numberGiven

1)+ x3x2 7x5( 3+ x3x2 7x5 ×= 3

3. Can the number 6n , n is a natural number end

with the digit 5? Give reason for your answer.

5. digit thewith end cannot 6 Hence,

odd. always is 5 is digit last whose number Any

even. always is6 So,

6 of factor a is 2

3 23)2(6

n

n

n

nnnn ×=×=

4. For what values of natural number n, 4n can end

with the digit 6?

6.with ends 4even, isn When

4096= then 4 6, =n if

1024 = then 4 5, =n if

256 = then 4 4,=n if

64 = then 4 3, =n if

16 = then 4 2, =n if

4= then 4 1, =n if

n

6

5

4

3

2

1

∴

5. Find the HCF of 252525 and 363636.

10101 =

37 137 3 = H.C.F

37137 3 2 = 363636

7 3 137 3 5 = 25252532

2

×××

××××

××××

6. If 13824 = 2a ×3b then find a and b.

3. b ,9 a Hence

3 x 2 = 13824 39

==

7. a,b are two positive integers such that ab ×ba=

800. Find ‘a’ and ‘b’.

5b2, a thus

b× a

5×5×2 ×2×2×2×2

800 =b× a

ab

ab

==

×=

=

25 52

8. If m, n are natural numbers, for what values of

m, does 2n x 5m ends in 5?

• For any value of n, 2n will become even.

• For any value m, 5m ends with 5.

• Product of even number and a number ends

with 5, we get a number ends with 0.

• We should not apply 0 for n and m, because

n and m are natural numbers.

9. If 113400 1 p ,p ,p ,p 4321 x

4x

3x

2x

1 = find the

value of 4321 p ,p ,p ,p ( primes in ascending

order) 432 x ,x ,x ,x1 ( integers, )

1 ,2 4,3, x ,x ,x ,x

7 5, 3, 2, p ,p ,p ,p

7 x 5 x 3 x 2 = 113400

4321

4321

1243

=

= .

10. Find the greatest number consisting of 6 digits

which is exactly divisible by 24,15,36?

99972036 and 24,15 by divisible

number digit-6 greatest

999720. = 279 - 999999 i.e

279. remainder i.e

360

999999.

36 ,24,15 of LCM

number digit-6 Greatest

36036 ,24,15 of LCM

999999. number digit-6 Greatest

=

=

+×=

=

=

=

2793602777

11. Find LCM and HCF of 408 and 170 by applying

the fundamental theorem of arithmetic.

2040 = L.C.M

17 x 5 x 3 x 23 = L.C.M

34 = H.C.F

17 ,2 factorsCommon

17 x 5 x 2 = 170

17 x 3 x 23 = 408

=

12. What is the smallest number that when divided

by three numbers such as 35, 56 and 91 leaves

remainder 7 in each case?

3647 =number Smallest Hence,

7 + 3640 = caseeach in 7 remainder a leaves

3640 =

x13 x2 2 x x2 5 7x = LCM

13 x7 = 91

7 x 2 x 2 x 2 = 56

7 x 5 = 35

,9156 35, of LCM = 91 56, 35, by divided no. Smallest

13. Find the least number that is divisible by the

first ten natural numbers.

2520 number smallest The

2520 =

7 × 5 × 3 × 3 × 2 × 2 × 2 numbers these of L.C.M

10 9, 8, 7, 6, 5, 4,3, 2, 1, :numbers natural 10

=

=


7. MODULAR ARITHMETIC

1. Find the least positive value of x such that

(i) 71 ≡ x (mod 8)

7x of valueleast

8 by divisible is 8

7 - 71 n

8

x - 71 n

8n = x - 71

=

=

=

(ii) 78 + x ≡ 3 (mod 5)

0x of valueleast

5 by divisible is,5

075

5

x75n

5n = x + 75

5n = 3 - x + 78

=

+=

+=

(iii) 89 ≡ (x + 3) (mod 4)

2x of valueleast

4by divisible is2 - 86

n

x - 86

n

4n= 3 - x - 89

=

=

=

4

4

(iv) 5) (mod 7

x 96 ≡

7 x of valueleast

7

x-96

7

x-96

=

=

=

5

5

n

n

.

(v) 5x ≡ 4 (mod 6)

2=

=

=

x of valueleast

6 by eisdivisibl6

4- 5(2)n

6

4- 5xn

6n = 4- 5x

vi) 67 + x≡1 (mod 4)

2x of valueleast

4by divsible is4

2674

x66n

4n x + 66

4n 1-x +67

4)(mod 1x +67

=

+=

+=

=

=

≡

vii) 98≡ (x+4) (mod 5)

4x of valueleast

5 by divsible is4

4675

x9n

5n x - 94

5n4-x - 98

=

−=

−=

=

=

4

2. Solve the following congruent equations

i) 5x ≡ 4 (mod 6)

.2,8,14,...

6,.......62 6,2 2, x of values

2 x of valuesleast

6 by divisible is6

4- 5(2) =n

6

4- 5x =n

6n = 4- 5x

=

+++=

=

ii) 3x −2 ≡ 0 (mod 11)

8,19,30

.11),......11(8 11),(8 8, x of values

8 x of valueLeast


2 - 3(8) =n

11

2 - 3x =n

11n = 2 - 3x

=

+++=

=

iii) 8x ≡ 1 (mod 11)

..7,18,29,..

.11),......11(7 11),(7 7, x of values

7 x of valueLeast


1 - 8(7) =n

11

1 - 8x =n

11n = 1 - 8x

=

+++=

=

iv) 3x ≡ 1 (mod 15)

solution integer no is there So,

integeran be tcan' 15

1-

5

x =n

15

1 - 3x =n

15n = 1 - 3x

3. Compute x, such that 104 ≡ x (mod 19)

( )

6

19610

192510

19510

19510

4

4

222

2

=

≡

≡

≡

≡

xThus

)(mod

)(mod

)(mod

)(mod


4. Prove that 2n + 6×9n is always divisible by 7 for any positive integer n.

7 by divisible is it Hence

)(

)(2

63m+9)- 2 ( 2

9 2-63m+ 2 2

9 )2-(7m+ 2 2 =

9 9× 6 + 2 2 =

9×6 + 2 = 1)P(k

1 +k =n for true is p(x) : prove To

2-7m = 9×6

7m = 9×6 + 2

7. by divisible is 9×6 + 2 =p(k)

k =n for true is p(x):Assume

7. by divisible is 56

56 = 54 + 2 =

9×6 + 2 =

1 =n for true is p(x) : prove To

k

k

kk

kk

kk

1+k1+k

kk

kk

kk

11

km

m

297

797

−=

×+−=

=

××=

××

××

+

5. Find remainders if70004 and 778 is divided by 7.

1. Remainder

7) (mod 1 778

7) 1(mod 1777

7) (mod 0 777

:7 by divisible is777 ii)

4. Remainder

7) (mod 4 70004

7) (mod 40 470000

7) (mod 0 70000

:7 by divided is 70004i)

=

≡

+≡+

≡

=

≡

+≡+

≡

0

6. Find the remainder when 281 is divided by 17.

2 =

2 1 =2 (-1) = Thus,

-1(mod17)16 but,

2 (16) =

2)(2 =

2 2 =2

)(mod

20

20

204

18081

××

≡

×

×

×

≡

81

81

2

1702

7. .Flight travel from Chennai to London through British Airlines is 11 hours. The airplane begins its journey on Sunday at 23:30 hours. If the time at Chennai is 41/2 hours ahead to that of London’s time, then find the time at London, when will the flight lands at London Airport. As journey time is 11 hrs, flight will reach London airport at 23:30 + 11 hrs = 10:30 Chennai time.

Chennai time is 4:30 hrs ahead of London time, so at the time of landing time will be 10:30 - 4:30 hrs = 6:30 London time.

8. Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?

Associate 0, 1, 2, 3, 4, 5, 6 to represent the weekdays from Sunday to Saturday

Friday. 5 number for Day

7) (mod 5

7) (mod 5 7) (mod47

after days 45 come will uncle

2. Tuesday for Number

=

≡

≡

=

=

9. Today is Monday . Vani celebrated her birthday 75

days ago Find when Vani celebrated her birthday.

Associate 0, 1, 2, 3, 4, 5, 6 to represent the weekdays from Sunday to Saturday

Wednesday. birthday VanisThus,

Wednesday. 3 number for Day

7) (mod 3 75-1 Thus,

7) (mod 3

7) (mod 4-7

7) (mod 4- 7) (mod 74-

ago days 75birthday Vanis

1. Monday for Number

=

=

≡

≡

≡

≡

=

=

10. A man starts his journey from Chennai to Delhi by

train. He starts at 22.30 hours on Wednesday. If it

takes 32 hours of travelling time and assuming that

the train is not late, when will he reach Delhi?

hrs 6.30 at Fridayon Reach

Friday Thursday

8 24)(1 32

(mod24) 6.30

.(mod24) 54.30 24) (mod 3222.30

:timeReaching

24. modulo use we Here

hours. 32 time Travelling

22.30 timeStarting

+×=

≡

≡+

=

=

11. i)What is the time 100 hours after 7 a.m.?

am 11 47 24) (mod 1.7

4Reminder

4 24)(4 100

24) (mod 1.7:24 modulousing

=+=+

=

+×=

+

0000

0000

ii) What is the time 15 hours before 11 p.m.?

am 8

3-1112) (mod 1.1

Reminder

3 )1(1 1

12) (mod 1.1:12 modulousing

=

=−

=

+×=

−

5001

3

25

5001


8.Sequences

1. Find next three terms of the following sequence. (i) 8, 24, 72, …

1944 648, 216,: terms 3 Next

1944 = 648(3) = a

648 = (3) 216 = a

216 = 72(3) = a

3. by multiplied is termEach

6

5

4

.

(ii) 5, 1,-3,…

15- = 4- 11- = a

11- = 4-7 - = a

7- = 4- 3- = a

4.by decreases termEach

6

5

4

iii) ,...,,14

1

6

1

2

1

26

1

422

122

1

418

118

1

414

1

6

5

=+

=

=+

=

=+

=

a

a

4a

4.by increased is rdenominato

same are numerators

iv) 4125 −− ,,,

13310

1037

734

7

6

−=−−=

−=−−=

−=−−=

a

a5a

3. by decreased is termEach

v) ,.....,., 010101

00001010

00010

0001010

0010

001010

010

..

a

..

a

..

a

10. by divided is termEach

6

5

4

==

==

==

vi) 1/4, 2/9, 3/16,…

......,36

5, : terms 3 Next

)(

6 = a

36

5

)(

5 = a

)(

4 = a

)(

n = a

6

5

4

n

49

6

25

4

49

6

16

15

25

4

14

1

2

2

2

2

=+

=+

=+

+n

2. Find the first four terms of the sequences whose nth terms are given by

(i) an = n3 −2

62. 25, 6, 1,- : terms four 1st

62 = 2- 4= a

25 2-3 = a

6 = 2-2 = a

1 = 2-1 = a

2-n = a

34

33

32

31

3n

=

(ii) an = (−1)n+1 n(n + 1)

20.- 12, 6,- 2, : terms four 1st

-201) + 4(4(-1) = a

121) + 3(3 (-1) = a

-61) + 2(2 (-1) = a

2 1) + 1(1 (-1) = a

1) +n(n (-1) = a

1+44

1+33

1+22

11+1

1+nn

=

=

=

=

(iii) an = 2n2 - 6

26 12, 2, 4,- : terms four 1st

266 - 2(4) = a

126 - 2(3) = a

26 - 2(2) = a

-46 - 2(1) = a

6 - 2n = a

24

23

22

21

2n

=

=

=

=

3. Find nth term /general term for the sequences i) 3, 6, 9

n3=na

3. of multiples are terms

ii)2, 5, 10, 17,...

1 + n = a

1),... + (4 1), + (3 1), + (2 1), + (1

1),... + (16 1), + (9 1), + (4 1), + (1

2n

2222

iii) ,.....,2

1

3

2..........

1

13

3

12

2

11

1

+=

+++=

n

nna

Nrthan more 1 is Dr n, is termnth of Nr.

,,

iv) ,.....,2

1 0,

3

2..........

n

1 -n a

,.....,2

1-2 ,

1

1-1 =

r.denominatothan less 1 is Numerator

n, is termnth of r.Denominato

n =

−

3

13

v) 3, 8, 13, 18,...

2 -5n a

2),... - (20 2), - (15 2), - (10 2), - (5 =

18,... 13, 8, 3, =

n =


vi) 5,-25,125,……

5 )(a

5. of powersin are they

ealternativsign - and

.. ,5 ,5- ,5 =

nn

321

11 +−=

+

n

4. Find the indicated terms of the sequences whose nth terms are given by

(i) an = 5n / (n + 2); a6 and a13

3

138

30

=

=2 + 13

5(13) = a

4

15 =

2 + 6

5(6) = a

2 +n

5n = a

136

n

(ii) an = -(n2 - 4); a4 and a11

117- 4)- (11- = a

12- 4)- (4- = a

4)- (n- = a

211

24

2n

=

=

5. Find a11 and a18 whose general term is

∈+

∈+=

Nnodd, is);(

Nneven, is;

nnn

nnan

3

12

154

31111

3

325

118

12

2

=

+=

+=

=

+=

+=

)(a

odd )(

a

even)(

1118

nnnannann

6. Find a8 and a15 whose nth term is

∈+

∈+

−

=

Nnodd, is;

Nneven, is;

nn

n

nn

n

an

12

3

1

2

2

31

225

1152

15

12

11

6338

18

3

1

2

2

2

2

=

+=

+=

=

+

−=

+

−=

)(a

odd is;

a

even is if

158

nn

nan

n

na nn

7. If a1 = 1, a2 = 1 and an = 2an - 1 + an - 2 n ≥ 3, n ∈ N, then find the first six terms of the sequence.

17 4

72(17)

a + 2a

a + 2a = a

1

32(7)

a + 2a

a + 2a = a

7

12(3)

a + 2a

a + 2a = a

3

12(1)

a + 2a

a + 2a = a

a + 2a = a

1 = a 1, = a

45

2 - 61 - 66

34

2 -51 - 55

23

2 - 41 - 44

12

2 - 31 - 33

2 -n 1 -n n

21

=

+=

=

=

+=

=

=

+=

=

=

+=

=

8. a1 = 1, a2 = 1 fin first 6 terms of

3≥+

n ,3 a

a = a

2 -n

1 -n n

52

1

34

116

1

3

16

131

3

4

131

1

3

3

3

4

41

2

3

1

2

=

+

=+

=

+

=

+=

+=

+

=

+=

+=

+

≥+

a

a

a

a

a

a

3 a

a = a

3 a

a = a

3 a

a = a

n ,3 a

a = a

1 = a 1, = a

2 - 5

1 - 55

2 - 4

1 - 44

2 - 3

1 - 33

2 -n

1 -n n

21

9. Arithmetic Progression

1. Check whether following sequences are in A.P. i) a - 3, a - 5, a -7,...

A.Pin is sequencegiven

t - tt - t

2- =

5 + a -7 - a =

5) - (a - 7) - (a =t - t

2- =

3 + a - 5 - a =

3) - (a - 5) - (a = t - t

2312

23

12

∴

=

=

ii) 1/2, 1/3, 1/4, 1/5........

A.Pin not is sequenceGiven

t - tt - t

t - t

6

1 =

3

1 = t - t

2312

2312

∴

≠

−=

−=−

12

13

1

4

1

12

1

iii) 9, 13, 17, 21, 25,...

A.Pin is sequenceGiven

t - tt - t

13-7 t - t

4=

9-3 = t - t

2312

23

12

∴

=

=

=

4

1

1

iv) -1/3, 0, 1/3, 2/3.........



t - tt - t

3

1 t - t

3

1 =

3

1-- = t - t

2312

2312

∴

=

=

−=

3

1

00

v) 1, -1, 1, -1, 1, -1,............


t - tt - t

)(1 t - t

2- = 1-1 = t - t

2312

23

12

∴

≠

=−−=

−

21

vi) ,....,, 43322 +++ xxx


t - tt - t

)()( t - t

)()( = t - t

2312

23

12

∴

=

−=+−+=

−=+−+

13243

1232

xxx

xxx

vii) 2, 4, 8, 16,…..


t - tt - t

4- t - t

2 = 2- = t - t

2312

23

12

∴

≠

== 48

4

viii) ,....,,, 29272523


t - tt - t

t - t

= - = t - t

2312

23

12

∴

=

=−= 222527

222325

2. Write an A.P. whose first term is 20 and

common difference is 8.

44,36, 28, 20,

,...838,202 20 ,820 20,

3d,... + a 2d,+ a ,da a,

:n Progressio Arithmetic

8 =d ,differencecommon

20 a term, First

×+×++=

+=

=

3. Find first term and common difference of Arithmetic Progressions whose nth terms are given (i) tn = −3 +2n

2differencecommon

1- term First

2 = d

(-1) - 1

t - td

2(2) + 3 t

1- = 2(1) + 3- = t

2n 3-t

12

2

1

n

=

=

=

=

=−=

+=

1

(ii) tn = 4 - 7n

7-differencecommon

1- term First

-

(-3) - 10- =d

7(2) - 4t

3- =

7(1) - 4= t

7n - 4t

2

1

n

=

=

=

=

−=

=

=

7

10

4. Find 15th , 24th and nth term (general term) of

an A.P. given by 3, 15, 27, 39,…

912

12123

1213

279

12243

121243

171

12143

121153

1

24

24

15

15

−=

−+=

−+=

=

×+=

−+=

=

×+=

−+=

−+=

==

=

nt

n

nt

t

t

t

t

dna

n

n )(term, n

)(

)(

)(t, term General

. 123- 15 d ,difference com.

3 a ,term first

th

n

5. Find the 19th term of an A.P. -11,-15,-19,...

83- = t

72 - 11 - =

18(-4) + 11- =

1)(-4) - (19 + 11- = t

1)d -(n + a = t term n

4- = 11 + 15- =

(-11) - 15- = d

,11- = a

19

19

nth

6. Determine the general term of an A.P. whose 7th

term is −1 and 16th term is 17

152

2213

2113

13126

218921

21715

17116

17

116

117

1

1

167

−=

−+−=

−+−=

−=⇒−=+

=

=⇒−=−−

=+

=−+

=

−=+

−=−+

−=

−+=

n

n

n

aa

dd

da

da

t

da

da

t

dna

n

n

nth

t

)(t

)(

(1)in 2 d Put

)()(

)......(

)(

)......(

)(

)(t, term n

7. Find number of terms in 3, 6, 9, 12,…, 111


37 =n terms, of No.

1 + 36 =

1 +3

3 - 111 =

= terms, of No.

33 - 6 = .difference com.

111 = term last

3 = a term First

+

−

=

1d

aln

l

8. Which term of an A.P. 16, 11, 6, 1,... is -54 ?

15 =n terms, of No.

1 + 14 =

1 +5-

16 - 54- =

= terms, of No.

-16 - 11 = .difference com.

54- = term last

16 = a term First

+

−

=

1

5

d

aln

l

9. Find middle terms of A.P. 9, 15, 21, 27,…,183.

99 =

15(6) + 9 =

15d + a = term16th

93 =

14(6) + 9 =

14d + a = term15th

term 2

nand term

2

n : terms middle

30(even) =n

1 +6

9 - 183 =n

=n terms of No.

69 - 15 = ddifference com.

183 = lterm last

9, = a term First

thth

+

+

−

=

1

1d

al

10. If nine times ninth term is equal to the fifteen times fifteenth term, show that six times twenty fourth term is zero.

0 = t

0 = 23d + a

0 = 23d) + 6(a

0 = 138d + 6a

0 = 72d - 210d + 9a - 15a

d 210 + 15a = 72d + 9a

14d) + 15(a = 8d) + 9(a

15t = 9t

24

159

.

11. If 3 + k, 18 - k, 5k + 1 are in A.P. then find k

4=k

16 =4k

k +3k = 2 - 18

2 +3k =k - 18

4+6k = k) - 2(18

1 +5k +k + 3 = k) - 2(18

c + a = 2b

:A.Pin are c b, a,

1 +5k = c k , - 18 = b k, + 3 = a

12. Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.

3 = x20 =17 + x

20 = y+ xin yApplying

31 = z 48= z +17

48= z + y

24 + 24 = z + y

24 - z = y- 24

t - tt - t

17 = y34 = 2y

10 + 24 = y+ y

y- 24 = 10 - y

t - tt - t

(1)--- 20 = y+ x

10 - y x - 10

t - t= t - t

:A.Pin is sequenceGiven

4534

3423

2312

⇒

⇒

=

⇒

=

=

Hence the values of x, y and z are 3, 17, 31 respectively.

13. In a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?

78. row lastin seats of No.

78 = l

2 1) - (3020 -

t

30 = (n) rows of No. Total

30 =n and 2 = d 20, = a

24 = 2 + 22 = row 3in seats of No.

22 = 2 + 20 = row 2in seats of No.

20 = row 1in seats of No.

n

rd

nd

st

=

=

=+

−

=+

−

=

l

l

d

al

3012

20

301

30

14. In an A.P., sum of 4 consecutive terms is 28

sum of their squares is 276. Find the 4 numbers.


2 ± = d )(

4a

3d) + (ad)a (d)-a (3d)-(a

276, squares their of Ssum

7 = a 28 4a

28 3d + a da d-a3d-a

28, terms 4of Sum

terms Four3d + a d,a ,d-a 3d,-a

2

2222

⇒=+

=+

=

++++++

−++−+

=++++

=

⇒=

=++++

=

→+

2762074

27620

276269

269

276

22

2

2222

2222

d

d

addaadda

addaadda

13 and 9 5, 1,

3(2)+7 ,2 ,72-7 3(2),-7 : Numbers

2 = d7,a

+

=

15. Sum of three consecutive terms that are in A.P. is 27 Their product is 288. Find the three terms.

16. 9, 2, terms three First

16 =7 + 9 = d + a = term 3rd

9 = a = term 2nd

2 =7 - 9 = d - a = term 1st

7 = d 49- = d -

81 - 32 = d -

32 = )d - (9

288 = )d - 9(9

288 = )d - a(a

288 = d) + (a a d) - (a

288 = terms 3 of Product

9 = a27 = 3a

27 = d + a + a + d - a

27 = terms 3 of Sum

d. + a a, d, - a terms 3

2

2

22

22

22

=

⇒

⇒

=

16. A mother devides 207 into three parts such that the amount are in A.P. and gives it to her three children. The product of the two least amounts

that the children had 4623. Find the amount received by each child.

,71.69 67,

2+69 69, 2,-69 :given Amount

2 = d

4623d)69-(69

4623= )(

4623. amounts least 2 of product

69 = a

207 3a

207 =

207amount the of sum Since,

,, :children 3 by received Amount

=

−

=

=

+++−

=

+−

ada

daada

daada

17. The ratio of 6th and 8th term of an A.P. is 7:9. Find the ratio of 9th term to 13th term.

7. : 5 term 13 : term 9

14d

10d =

12d + 2d

8d + 2d =

12d + a

8d + a =

t

t

term 13 : term 9

2d = a

0 = 2d - a

0 = 4d- 2a

0 = 49d- 45d+ 7a - 9a

49d+ 7a = 45d+ 9a

7d) + 7(a = 5d) + 9(a9

7 =

7d + a

5d + a

9

7

t

t

thth

13

9

thth

8

6

=

=

18. Temperature of Ooty from Monday to Friday to

be in A.P. Sum of temperatures from Monday to

Wednesday is 0° C. Sum of the temperatures

from Wednesday to Friday is 18° C. Find the

temperature on each of the five days.

C 9° = 2(3) + 3 = 2d + a

C 6° = 3 + 3 = d + a

C 3° = a

C 0° = 3 - 3 = d - a

C 3°- = 2(3) - 3 = 2d - a

eTemperatur

Friday

Thursday

Wednesday

Tuesday

Monday

Days

3 = d

3 - 6 = d

6 = d + 3 (1),in a of valueApply

3 = a

6 = 2a

6 = a + a (2) + (1)

(2)--- 6 = d + a

18 = 3d + 3a

18 = 2d + a + d + a + a

1friday to Wednesday from sum

(1)--- 0 = d - a

0 = 3d - 3a

0 = a + d - a + 2d - a

0Wednesday to Monday from sum

2d + a ,d + a a, d, - a 2d, - a

: Friday to Monday from temp.

8=

=


19. Priya earned 15,000 in the first month.

Thereafter her salary increased by 1500 per

year. Her expenses are 13,000 during the first

year and the expenses increases by 900 per

year. How long will it take for her to save

20,000 per month.

years31in 20000 save will she

30.8 = 12

361 yearsof number

361 =n months, of number

=

=n months, of number

-52000-2050d

200003000, a

APan is It

0002100,..,20 2050, 2000,:montheach ofSaving

13150,.... 13075, 13000,:sequence a as Expenses

75 = 12

900 = increment monthly

900 = earexpenses/y increasing

13000 = expensemonth 1st

15250,.. 15125, 15000, :sequence a as Earnings

125 = 12

1500 = increment monthly

1500 = year/ salary of Increment

=

=

+

−

+

−

==

==

150

200020000

1d

al

l

20. If lth, mth and nth terms of an A.P. are x, y, z

respectively, then show that

i) 0=−+−+− )()()( mlzlnynmx

0

00

111

1

1

111

=

+=

−−+−−+−+−−+

−+−+−=

−−++

−−++−−+=

−+−+−

=−+

=

=−+

=

=−+

=

)()(

))(())(()())([(

)]()()[(

)]()([

))](([)]()([

)()()(

)()()(

da

nmlmlnnmlnmd

lnnmmla

mldna

lnlnyanmdla

mlzlnynmx

zdna

zt

ydma

yt

xdla

xt nml

ii) 0=−+−+− mxzlzynyx )()()(

0=

−+−+−=

−+−+−=

−+−+−

−=−

−=−

−=−

)(

])()()[(

)()()(

)(

)(

)(

lmnmnl lmmnlnd

mlnlnmnmld

mxzlzynyx

dlnxz

dnmzy

dmlyx

10. Arithmatic Series

1. Find the sum of the following (i) 3, 7, 11,… up to 40 terms.

3240 =

156] + [6 20 =

1)4] - (40 + [2(3) 2

40 =

1)d] -(n + [2a 2

n = S

4= 3 -7 = (d) differenceCommon

3 = (a) term First

40= (n) terms of Number

n

(ii) 102, 97, 92,… up to 27 terms.

999 =

(74) 2

27 =

130] - [204 2

27 =

26(-5)] + [204 2

27 =

1)(-5)] -(27 + [2(102) 2

27 =

1)d] -(n + [2a =Sn

5- = 102 -97 = (d) differenceCommon

102 = (a) term First

27 = (n) terms of Number

2

n

(iii) 6 + 13 + 20 +...........+ 97

721 =

7[103] =

97] + [62

14 =

][ 2

n =Sn

14 = 1 +7

6-97 =

1 + =n terms of No.

7 = 6 - 13 = d 6, = a 97, = l

a + l

d

al

−

.

iv) 1460430400 ++++ ........ .

14.7 =

1] + [0.402

21 =

][ 2

n =Sn

21 = 1 +0.03

0.4-1 =

1 + =n terms of No.

7 = 6 - 13 = d 6, = a 97, = l

a + l

d

al

−


vi) ,...,,,4

35

2

16

4

178 to 15 terms

4

165

2

2116

2

15

4

311582

2

15

122

4

38

15

15

=

−=

−−+=

−+=

−=−

S

S

dnan

S

d

n

)()(

])([

4

17 =,differenceCommon

8 = a term First

15 =n terms, of Number

2. How many consecutive odd integers beginning with 5 will sum to 480?

20 =n or 24- =n

0 = 20) -(n 24) +(n

0 = 480- 4n + n)(

960 =8n + 2n

480= 8] +[2n

480= 2] -2n + [10

480= 1)(2)] -(n + [2(5)

480= 1)d] -(n + [2a

480=Sn

2 = 5 -7 = d 5, = a

480= ............. + 11 + 9 +7 + 5

2

2

2

2

2

2

2

÷

n

n

n

n

3. How many terms of the series 1+5 +9 +... must

be taken so that their sum is 190?

×

+

2

19- =n or 10=n

0 = 19)(2n 10)-(n

0 = 190 -n -n

380 = 2]-n[4n

190 = 4]- 4n +[2

190 = 1)(4)] -(n + [2(1)

190 = 1)d] -(n + [2a

190 =Sn

4= 2 - 5 = d 1, = a

190 = ...........+ 9 + 5 + 1

22

2

2

2

n

n

n

4. Find the sum of first 28 terms of an A.P. whose nth term is 4n -3.

1540 =

[110] 14 =

27(4)] + 14[2 =

1)4] - (28 + [2(1) 2

28 = S

1)d] -(n + [2a = S

4= 1 - 5 = d

5 =

3 - 4(2)= t

1 =a

3 - 4=

3 - 4(1)= t

3 - 4n =tn

28

n

21

2

n

5. The sum of first n terms of a certain series is given as 2n2 -3n . Show that the series is an A.P.

4is d whose A.Pan is It

7,.... 3, 1,-:series

t2

9 =ttt

9 = 3(3) - 2(3) = S

t1-

2 =tt i.e,

2 = 3(2) - 2(2) = S

1- = t i.e,

3 - 2 = 3(1) - 2(1) = S

3n - 2n = S

3

321

23

2

21

22

1

21

2n

729

9

312

2

3

2

=−=

=+

++

=+=

=+

+

t

t

6. In an A.P. the sum of first n terms is 2

3

2

5 2 nn+ .

Find the 17th term.

48466-748

SST

2

1280

)()( = S

S2

1445

)()( = S

= S

161717

16

17

17

n

==

−=

=

+=

+

=

+=

+

+

6642

482

163

2

165

7482

512

173

2

175

2

3

2

5

2

2

2 nn

7. The 104th term and 4th term of an A.P. are 125 and 0. Find the sum of first 35 terms.


612.5 = S

2

35 =

4

51) - (35 +

4

15-2

2

35 =

1)d] -(n + [2a 2

n =Sn

4

15- = a 0 =

4

53 + a (2),in sub

4

5 =d

100

125 = d

125 = 100d (2), - (1)

(2)----- 0 = 3d + a

0 = t

(1)----- 125 = 103d + a

125 = t

35

4104

⇒

⇒

2

70

d

8. Find sum of all odd positive integers less than 450

50625 = S

4502

225 =

449]+ [12

225 =Sn

l] + [a2

n =Sn

225 =n

1 + 24 =

1 + 2

1-449 =n

1 + =n

2, = 1 - 3 = d 449,= l 1, = a

449+ 447 + .............+7 + 5 + 3 + 1

225

×

2

d

l-a

9. Find the sum of all natural numbers between 602 and 902 which are not divisible by 4.

224848 =

299(752) =

[1504]2

299 =

901] + [603 2

299 = S

l] + [a2

n =Sn

299 =n

1 + 298 =n

1+1

603-901 =n

1 +d

a-l =n

901 + ........+ 604 + 603

902 and 602 b/w no. of Sum

301

902 and 602 b/w 4by divisible no. of Sum

56400 =

75(752) =

[1504]2

75 =

900] + [604 2

75 = S75

l] + [a2

n =Sn

75 =n

1 + 4

604 - 900 =

d

a-l =n

900 + ..... + 612 + 608 + 604 =

+

1

168448 =

56400 - 224848 =

902 &602 b/w

4by divisible

no. of Sum

-

902 and 602

between

no. of Sum

=

902 and 602 b/w

by4 divisible

not no. of Sum

10. Find the sum of all natural numbers between

300 and 600 which are divisible by 7.

19264 =

595] + [301 2

43 = S

l] + [a2

n =Sn

43=n

1+7

301-595 =n

1 +d

a-l =n

595. ,… 315, 308, 301,

are7 by Divisible

2,.....600300,301,30

:600 and 300 b/w numbers Natural

301

11. Raghu can buy a laptop by paying 40,000 cash or by giving it in 10 installments as 4800 in the first month, 4750 in the second month, 4700 in the third month and so on. Find (i) total amount paid in 10 installments. (ii) how much extra amount that he has to pay than the cost?

5750 =

40000- 45750= amount Extra

45750=

9(-50)] + 5[9600 =

(-50)] 1) - (10 + [2(4800)2

10 = S

d] 1) -(n + [2a2

n = S

50- = 4800- 4750= d 4800,= a 10, =n

tsinstallmen 10 ...to+ 4700+ 4750+ 4800

10

n

12. A man repays a loan of 65,000 by paying 400 in the first month and then increasing the payment by 300 every month. How long will it take for him to clear the loan?


months 20 :in clear amountloan

(x) 3

65- =n 20, =n 0 = 65) +(3n 20) -(n

0 = 20) -65(n + 20) -3n(n

0 = 1300 -65n +60n - 3n

0 = 1300 -5n + 3n

1300 = 3n2 +5n

1300 = 3n] + n[5

650 = 3n] + [5

65000 = 300] -300n + [800

65000 = 1)300] -(n + [2(400)

65000 = 1)d] -(n + [2a

65000 = amountloan

................ 1000, + 700 + 400

1000 = 300 + 700 =month 3rd For

700 = 300 + 400=month 2nd For

400=month 1st For

2

2

⇒

2

2

2

2

n

n

n

n

13. A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two bricks less than the previous step. How many bricks are required for top most step? How many bricks are required to build stair case?

2130 = 15(142) =

42]+ [1002

n =

l] + [a2

n =Sn (ii)

42= 58 - 100 =

29(-2) + 100 =

29d + a = t (i)

30 = (n) steps of number

2- = 100 - 98 = d

98 = 2 - 100 = step 2ndin bricks of No.

100 = (a) step 1stin bricks of No.

30

14. Find sum terms 12 ....to++

−+

+

−+

+

−

ba

ba

ba

ba

ba

ba 3523

+

−=

+

−−+

+

−=

−+=

+

−=

+

−−

+

−=

+

−=

ba

ba

ba

ba

ba

baS

dnAn

S

ba

ba

ba

ba

ba

baba

ba

n

13246

21122

2

12

122

223

12 )(

])([

enceCom.differ

A Term, First

15. If S1, S2, S3,....Sm are the sums of n terms of m A.P.’s whose first terms are 1,2, 3,...m and whose common differences are 1, 3, 5,..., (2m -1) Show

that 2

1321

)(....

mnmnSSSS m

+=+++

2

1

124

211222

12

2

1122

12151312

122

12122

12

152

51322

53

132

31222

32

12

11122

11

122

321

321

3

3

2

1

)(

][

])()([.....

Terms of No.

1) - 2n(m

1 )()(

Terms of No.

]...[....

)()])(()([

,)

)(])()([

,)

)(])()([

,)

)(])()([

,,)

])([

mnmn

mnmn

nmnmn

SSSS

m

n

n

nnmn

nmnnnnn

SSSS

nmnn

mnmn

S

mdmaiv

nn

nn

S

daiii

nn

nn

S

daii

nn

nn

s

dai

dnan

S

m

m

n

+=

+×=

−++=+++

=

+=

+

+−+−=

+−+++++=+++

+−=−−+=

−==

+=−+=

==

+=−+=

==

+=−+=

==

−+=

16. The sum of first n, 2n and 3n terms of an A.P.

are S1S2 , and S3. Prove that S3= 3(S2 – S1)

( )

312

12

12

3

2

1

3

1322

33

1322

1212242

122

1222

2

1322

3

1222

2

122

SSS

dnan

SS

dnan

dnadnan

dnan

dnan

SS

dnan

Sn

dnan

Sn

dnan

Sn

=−

−+=−

−+=

−+−−+=

−+−−+=−

−+=

−+=

−+=

)(

])([)(

])([

])([])([

])([])([

])([ terms 3 first of Sum

])([ terms 2 first of Sum

])([ terms first of Sum


11. Geometric Progression

1. Which of the following sequences are in G.P.? (i) 3, 9, 27, 81,…

GP a is It

t

t

t

t

t

t

t

t

2

3

1

2

2

3

1

2

∴

=

==== 39

273

3

9

(ii) 4,44,444,4444,...

GP not is It

t

t

t

t

44

444

t

t

t

t

2

3

1

2

2

3

1

2

∴

≠

=== 114

44

(iii) 0.5, 0.05, 0.005,…

GP a is It

t

t

t

t

0.1 =0.05

0.005

t

t0.1 =

0.5

0.05

t

t

2

3

1

2

2

3

1

2

∴

=

==

. c

(iv) 1/3, 1/6, 1/12,.......

GP a is It

t

t

t

t

t

t

t

t

2

3

1

2

2

3

1

2

∴

=

=×==×=2

1

1

6

12

1

2

1

1

3

6

1

(v) 1, −5, 25, −125,…

GP a is It

t

t

t

t

t

t

t

t

2

3

1

2

2

3

1

2

∴

=

=−

==−

= 55

255

1

5

(vi) 120,60,30,18,…

GP a is It

t

t

t

t

60

30

t

t

120

60

t

t

2

3

1

2

2

3

1

2

∴

=

====2

1

2

1

(vii) 16, 4, 1, 1/4,..........

GP a is It

t

t

t

t

t

t

16

4

t

t

2

3

1

2

2

3

1

2

∴

=

===4

1

4

1

viii) 7, 14, 21, 28, …

GP a not is It

t

t

t

t

t

t

7

14

t

t

2

3

1

2

2

3

1

2

∴

≠

====2

3

14

212

ix) ½,1,2 4, ...

GP a is It

t

t

t

t

t

t

2

1

1

t

t

2

3

1

2

2

3

1

2

∴

=

==== 21

22

x) 5, 25, 50, 75

GP not is It

t

t

t

t

t

t

5

25

t

t

2

3

1

2

2

3

1

2

∴

≠

=== 225

505

2. Write first three terms of the G.P. whose first term and the common ratio are given below. (i) a = 6, r = 3

54. 18, 6, : terms three 1st

54 = 6(3) = ar ,term Third

18 = 6(3) = ar ,term Second

6 = (a) term First

22

(ii) a = √2, r = √2

2,,2 : terms three First

22 = 222 = ar ,term Third

2 = 2 = ar term, Second

2 = (a) term First

2

22

(iii) a = 1000, r = 2/5

160. 400,1000, : terms three First

160 =5

2 1000 = ar term, Third

400= 5

2 1000 = ar term, Second

1000 = (a) term First

22

×

(iv) a =−7 , r = 6

252.42,--7,- : terms three First

252- =(6)7 - = ar term, Third

42- = 67- = ar term, Second

7- = a ,term First

22

×

(v) a = 256 , r = 0.5

,..6256,128, : terms three First

64 = 256 = ar term, Third

128 = 256 = ar term, Second

10

5r256, = a ,term First

22

4

2

1

2

12

1

×

==

3. i)In a G.P. 729, 243, 81,… find t7 . ii) Find 8th term of the G.P. 9, 3, 1,…


243

1 =

9 =

9 = t

ar =tn

9

3 = r

9, = a ii)

1 =

729 =

729 = t

ar =tn

729

243 = r

729, = a i)

7

1- 8

7

1-n

6

1-7

7

1-n

=

=

3

1

3

1

3

1

3

1

3

1

3

1

4. Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression.

18- = x

54- = 3x

144 - 90 = 21x - 24x

90 + 21x = 24x + 144

90 + 6x + 15x + x = 2x(12) + 122 + x

12 + x

15 + x

6 + x

12 + x

t

t

t

tG.P a For

22

2

3

1

2

=

=

.

5. Find the number of terms in the following G.P. (i) 4, 8, 16,…,8192 ?

12 =n

2 = 2

8192 = (2)

8192 = 4(2)

8192 = ar

8192 = t

2 = 4

8 = r 4,= a

111-n

1-n

1-n

1-n

n

(ii) 1/3, 1/9, 1/27,................1/2187

7 =n

2187

1 =

2187

1 =

3

1

2187

1 = ar

2187

19

1 = r

,3

1 = a

61-n

1-n

1-n

1-n

=

×

=

=×

3

1

3

1

33

1

3

1

3

1

1

3

nt

6. In a G.P. the 9th term is 32805 and 6th term is 1215. Find the 12th term.

1112

11

5

3 3

3

8

5

5

6

8

9

5(3) = t

ar term12th

5 = a343

1215 = a

1215 = a(3)(2),in r Apply

3 = r3 = r

27

1 =

r

1

32805

1215 =

ar

ar ,

(1)

(2)

-(2)- 1215 = ar

1215 = t

(1)- 32805 = ar

32805 = t

=

⇒

⇒

7. In a Geometric progression, the 4th term is 8/9

and the 7th term is 64/243 . Find the G.P.

,...,,

..3

2,

3

23,3 =

,...arar,a, GP

3 = a 8

27

9

8 = a

9

8 =

3

2a(1),in r Apply

3

2 = r

3

2 = r

27

8 = r

243

64 =

ar

ar ,

(1)

(2)

-(2)- 243

64 = ar

243

64 = t

(1)- 9

8 = ar

9

8 = t

2

3

3 3

3

3

6

6

7

3

4

3

423

3

8

9

2

=

=

⇒×

⇒

×

8. Find the 10th term of a G.P. whose 8th term is 768 and the common ratio is 2.

3072 =

6(512) =

)6(2 =

ar = t

6 = a128

768 = a

2] = r[768 = )a(2

768 = ar

768 = t

9

910

7

7

8

⇒

Q

9. If a, b, c are in A.P. then show that 3a, 3b, 3c are in G.P.


G.P.in are 3 ,3 ,3

2

ca = b

3 = 3

)(3 = 3

333

3

3 =

3

3 G.P,in are 3 ,3 ,3 If

t GP, For

2

cab

b-c = a-b A.P,in are c b, a, If

t AP, For

cba

c+a b

2

1c + a b

cab

b

c

a

bcba

2

2

∴

+

=

=

+=

−=−

2

2

3

1

231

t

t

t

ttt

.

10. If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb−c × yc−a × za−b = 1 .

( )

1

22

0

2222

2222

222

22

2

12

==

×=

×××=

××=

××=××

=⇒=

+=⇒+=

−+

−−+

−

−−−−

−−−

+−

−−

+−−−

o

caacacca

caacacca

caacca

caa

acc

cabaaccb

yx

zx

zzxx

zxzx

zyxzyx

xzyxzy

PGzyx

cabcab

)(

.in are,,

A.Pin are c b, a,

11. A man joined a company as Assistant Manager. The company gave him a starting salary of

60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?

76577 =

)60000(1.05 =

5%) + 60000(1 = years5 After

5%) + 60000(1 =

5%) + 5%)(1 + 60000(1 =

5%) + 60000(1 of 5% + 60000 = salary year3rd

5%) + 60000(1 =

60000 of 5% + 60000 = salary year2nd

60,000 = salaryStarting

5

5

2

12. Sivamani is attending an interview for a job and the company gave two offers to him. Offer A:

20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years. Offer

B: 22,000 to start with followed by guaranteed annual increase of 3% for first 5 years.

24040 =

)22000(1.03 =

3%) + 22000(1 = salary 4th year

3%) + 22000(1 =

22000 of 3% + 22000 = salary year2nd

22,000 = salaryStarting

5 =n and 0.03 = r 22000, = a,)

22820 =

)20000(1.06 =

6%) + 20000(1 = salary 4th year

6%) + 20000(1 =

20000 of 6% + 20000 = salary year2nd

20000 = salaryStarting

5 =n and 0.06 = r 20000, = a,)

3

3

3

3

BOffer

Offer Ai

ii

13. In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 57/2 . Find the three terms.

2

9 = 3 = ar ,term Third

3 = a ,term Second

2 = 3 = r

a ,term First

3

2 = r ,

2

3 = r0 = 2) - 3)(3r - (2r

0 = 6 + 13r - 6r

0 = 18 + 39r - 18r

0 = 18 + 57r - 18r + 18r

r57 = 18 + 18r + 18r

r57 = 1) + r + 18(r

2

57 =

r

)r + r + 9(1

2

57 = 1) + r +

r

1(a

2

57 = a + ra + a

2

57 = ar

r

aar .a .

r

a

2

57 = time a attaken terms 2 of Sum

3 = a27 = a

27 = ar a r

a


ar a, ,r

a : terms three

2

2

2

2

2

2

2

222

3

2

3

3

2

×

×

⇒

++

⇒

××

a


14. The product of three consecutive terms of a

Geometric Progression is 343 and their sum is

3

91. Find the three terms.

,,,: Numbers

,3

1(ii)Case

,,: Numbers

,(i)Case3

1 = r ,3 = r0 = 3) - 3)(r - (3r

0 = 3 + 10r - 3r

r 13 = 3 + 3r + 3r

r 13 = 1) + r + 3(r

3

91 =

r

)r + r + 7(1

3

91 = 1) + r +

r

1a(

3

91 = aa

r

a

3

91 = terms 3 of Sum

7 = a7 = a

343 = ar a r

a


ar a, ,r

a : terms three

2

2

2

2

33

3

7721

7

2173

7

73

==

==

⇒

++

⇒

××

ar

ar

r

15. The present value of a machine is 40,000 and

its value depreciates each year by 10%. Find the

estimated value of the machine in the 6th year

6023619

6

5

16

6

1

32

2

.6th yearin machine of Value

100

90 000 40

100

90 000 40

,100

90r

, 000 40 a GP, a is It

,.100

90 000 40,

100

90 000 40,

100

90 000 40i.e,

100

90 000 40 year2nd after machine of Value

100

90 000 40 yearfirst after machine of Value

=

×=

×=

=

==

=

×

××

×=

×=

−

−

t

art

n

nn

12. Geometric Series

1. Find the sum of first n terms of the G.P

(i) 5, -3, 9/5, -27/25, .........(ii) 256, 64, 16,…

−=

−

−

−

=

−=

−

−

−

<

n

n

n

n

rr

4

1

4

1

4

1

256 = S

r-1a = S

256

64 = r 256, = a ii)

5

3-

5

3-

5

3-

5 = S

r-1a = S

5

3- = r 5, = a i)

n

n

n

n

n

n

13

1024

1

1

1

4

1

18

25

1

1

1

1

2. i)Find sum of first 6 terms of G.P. 5, 15, 45, … ii) Find sum of 8 terms of the G.P. 1,-3,9, -27, ..

1640- =4

6561-1 =

(-3)-1

)(-3)-1(1 =S

)r-a(1 = S

1 3 - = 1

3- = r

1, = a 8, =n ii)

1820 =2

1) - 5(729 =

1 - 3

1) - 5(3 =S

)a(r = S

1 3 = 5

15 = r

5, = a 6, =n i)

8

8

n

n

6

6

n

n rr −

<

−

−

>

11

1

3. Find the first term of the G.P. whose i)common ratio 5. sum to first 6 terms is 46872.

ii) S6=4095 ,r = 4

3 = a

4095

34095a

340951) - a(4

40951 - 4

1) - a(4

4095= 1 - r

1) - (ra

4095= S ii)

12 = a

15624

446872a

46872(4)1) - a(5

468721 - 5

1) - a(5

46872= 1 - r

1) - (ra

46872= S i)

6

6

n

6

6

6

n

6

×=

×=

=

×=

=

=

4. Find the sum to infinity of (i) 9 + 3 + 1 + ..... (ii) 21 + 14 + 28/3 + .............

63 =

321 =

-1

21 =

r-1

a = S

3

2

21

14 = r21,= a ii)

27/2 =

39 =

-1

9 =

r-1

a = S

3

1

9

3 = r 9, = a i)

32

31

12×

∞

=

×

∞

=


5. i)First term of an infinite G.P. is 8 Sum to

infinity is 3

32. Find the common ratio.

ii)Find the least positive integer n such that 1+6 +62+....+ 6n >5000

6n Thus,

7 6

6

25001 6

55000 6

1-r

)a(r

S

6r1, = a ii)

4

1

4

3 - 1 = r

r - 1 = 4

3

r - 1 = 32

24

= r-1

8

= r-1

a

S i)

5

5

n

n

n

=

>

>

>

×>−

>−

=

=

=

=∞

776

46656

1

50001

5000

3

323

323

32

n

6. Find the sum of the Geometric series 3 + 6 + 12 +............+ 1536

3069 = S

3(1023) =

1) - 3(1024 =

1 - 2

1 - 23S

1 - r

1 - ra = S

10 =n

2 = 2

512 = 2

1536 = 3(2)

1536 =

1536 = t

2 = 3

6 = r3, = a

10

10

10

n

n

91-n

1-n

1-n

n

1-

=

nar

7. How many terms of 1+4+16 .. make the sum 1365

6

44

140956

=

=

+=

×=

=

=++

n

n

n

n

n

n

n

4

31365 1 - 4

13651 - 4

1) - 1(4

1365= 1 - r

1) - (ra

1365 = S

1365 n...... 16+4+1

8. Find rational form of i) 0.6666 ii) 1230.

333

41 =

0.999

0.123 =

0.123-1

0.123 =

r-1

a = S

.0.123

0.123r0.6,a

G.P a forms This

.. 0.01230.123 1230.

3

2 =

0.9

0.6 =

0.1-1

0.6 =

r-1

a = S

.0.6

0.06r0.6,a

G.P a forms This

006 0.0.060.6 0.6666

∞

===

++=

∞

===

+++=

1010

9. Find the sum to n terms of the series

(i) 0.4 + 0.44 + 0.444 +........ to n terms

×

9

(0.1) - 1 -n

9

4 =

0.1-1

(0.1) - 1 0.1 -n

9

4 =

terms]...n + 0.13+0.12+[0.1 - terms)1...n +1+[(19

4 =

terms].......n + 0.13) - (1 + 0.12) - (1 + 0.1) - [(19

4 =

terms]..n ..........+ 0.999 + 0.99 + [0.9 9

4 =

terms)n to ...... + 0.111 + 0.11 + (0.1 9

9 4=

terms)n to ...... + 0.111 + 0.11 + 4(0.1=

termsn to ........+ 0.444 + 0.44 + 0.4

n

n

(ii) 3 + 33 + 333 + ........... to n terms

×

3

n-

27

1) - 10(10 =

n-9

1) - 10(10

3

1 =

n-1-10

1) - 10(10

3

1 =

terms)]....n +1+1+(1-terms).....n + 1000+100+[(103

1=

terms]..n ..........+ 1) - (1000 + 1) - (100 + 1)-[(103

1 =

terms]..n ..........+ 999 + 99 + [99

3 =

terms)n to ...... + 111 + 11 + (1 9

9 3 =

terms]n +........... + 111 + 11 + [1 3 =

termsn to ........... + 333 + 33 + 3

n

n

n

(iii) 5 + 55 + 555 + ........... to n terms

×

9

5n-

81

1) - 50(10 =

n-9

1) - 10(10

9

5 =

n-1-10

1) - 10(10

9

5 =

term)]....n +1+(1-terms).....n + 1000+100+[(109

5=

terms]..n ..........+ 1) - (1000 + 1) - (100 + 1)-[(109

5 =

terms]..n ..........+ 999 + 99 + [99

5 =

terms)n to ...... + 111 + 11 + (1 9

9 5 =

terms]n +........... + 111 + 11 + [1 5 =

n to ........... + 555 + 55 + 5

n

n

n


10. A person saved money every year, half as much

as he could in the previous year. If he had

totally saved 7875 in 6 years then how much

did he save in the first year?

( )[ ]

40007875

1

1

=⇒=×

=−

=−

=

a

a

r

32

63a

7875-1

7875]r-a[1

.2

1 = r 6, =n

7875 Ssaved, amount Total

21

6

21

n

6

11. Kumar writes a letter to four of his friends. He asks each one of them to copy the letter and mail to 4 different persons with the instruction that they continue the process similarly. It costs 2 to mail one letter, find amount spent on postage when 8th set of letters is mailed.

174760 =

(87380) 2 = Cost

87380 =

3

65535 4=

]4[4 = S

]a[r =Sn

4.= 16/4 = r and 4= a 8, =n

series geometric a forms It

................... + 64 + 16 + 4

8

8

n

×

−

−

−

−

14

1

1

1

r

12. nyyxxyxyxyxyxSn ....)()()( +++++++++= 322322

prove that

−

−−

−

−=−

1

1

1

1 22

y

yy

x

xxSyx

nn

n

)()()(

1

1

1

1 22

432

432

443322

3223

22

3223

22

322322

−

−−

−

−=−

++

−++=

+−+−+−=

++++−

+++−++−=

++++

+++++−=−

+

++++++++=

y

yy

x

xxyxS

nyyy

nxxx

nyxyxyx

nyyxxyxyx

yxyxyxyxyx

nyyxxyx

yxyxyxyxyxS

n

yyxxyxyxyxyxS

nn

n

n

n

)()()(

)]term....(

term)....[(

terms....)()()(

]terms....))((

))(())((

]terms....)(

)())[(()(

y)- (x by sidesboth Multiply

....

)()()(

13.Special Series

1. Find the sum of the following series

i) 1 + 2 + 3 +............+ 60

1830 =

30(61) = 2

) 1 +60(60 = 60 +............+ 3 + 2 + 1

2

1) +n(n =n +...............+ 3 + 2 + 1

:numbers natural of Sum

ii) 1 + 2 + 3 +............+ 50

1275 =

25(51) = 2

) 1 +50(50 = 50 +............+ 3 + 2 + 1

2

1) +n(n =n +...............+ 3 + 2 + 1

iii) 2+4+6+......+80

1640 =

(41) (20) 2 =

2

1)+40(40 2 =

40)+............ + 3 + 2 + 2(1 =

iv) 3 + 6 + 9 + ............+ 96

1584 =

(33) (16) 3 =

2

1)+32(32 3 =

32) +............ + 3 + 2 + 3(1 =

v) 51+ 52 + 53 +............+ 92

3003 =

1275 - 4278=

25(51) - 46(93)=

21

1) + 50(50 -

2

1) + 92(92 =

50) +3... + 2 + (1 - 92) +..+ 3 + 2 + (1 = 92 +...+ 53 + 52 +51

vi)16+17+18+......+75

2730 =120 - 2850 =

15(8) - 75(38) =

21

1) + 15(15 -

2

1) + 75(75 =

15) +... 2 + (1 - 75) +..+ 2 + (1 = 75+......+17+16


i) 12+22+ 32+.......+192

2470 =

(13) 19(10) = 6

19(20)(39) =

6

1) + (2(19) 1) + 19(19 = squares of Sum


ii) 1 + 4 + 9 + 16 +...............+ 225

1240 =

(31) 5(8) = 6

15(16)(31) =

6

1) + (2(15) 1) + 15(15 =

15 +............+ 4+ 3 + 2 + 1 22222

iii)152+162+ 172+.......+282

82775 6

22125 =

6

1) + (2(21) 1) + 21(21 5 =

)21 +............+ 4+ 3 + 2 + (15

2

222222

=

×××

×

=

432

iv) 62 + 72 + 82 +........+212

3256 =

55 - 3311 =

6

(11) 5(6) -

6

(43) 21(22) =

6

1) + (2(5) 1) + 5(5 -

6

1) + (2(21) 1) + 21(21 =

6

1) +1)(2n +n(n = squares of Sum

)5 +.. + 2 + (1 - )21 +..... 2 +(1 = 21 +...+7 +6 22222 222 2

v)152+162+ 172+.......+282

6699 =

1015 - 7714 =

6

(29) 14(15) -

6

(57) 28(29) =

6

1) + (2(14) 1) + 14(14 -

6

1) + (2(28) 1) + 28(28 =

)14 +.. + 2 + (1 - )28 +..... 2 +(1 = 22222 2


i) 13 + 23 + 33 +.........+ 163

18496

6

=

=

+++

+++

2

2

333

2333

136

2

1) + 16(16 = 1..21

2

1) +n(n = n..21

ii) 103 + 113 + 123 +.........+ 203

42075=

(165) 255 =

45)- (210 45)+ (210 =

45- 210 =

2

1) + 9(9

2

1) + 20(20 =

)9 +...2 + (1 - )20 +....2(1 = 20 +... 11 + 10

22

22

333333333

−

+

ii) 93 + 103 + 113 +.........+ 213

52065

36) - (231 36) + (231 =

36 - 231 =

2

1) + 8(8

2

1) + 21(21 =

)8 +...2 + (1 - )21+....2(1 = 21 +... 10 + 9

22

22

333333333

=

−

+


i) 1 + 3 + 5 +.............+ 71

1296 =

36 =

2

1 =

2

1 = numbers odd of Sum

2

2

2

71

+

+l

ii) 1 +3+5+...... to 40 terms

784 =

28 =

2

1 = ...31

2

1 = ...31

2

2

2

55555

5

+++++

+++++

ll

iii)1 +3+5+...... +55

784 =

28 =

2

1 = ...31

2

1 = ...31

2

2

2

55555

5

+++++

+++++

ll

5. If 1+2+ 3 +......+n=666 then find n

, number, natural isn As

,

))((

n

6662

1)+n(n

666 =n +........... + 3 + 2 + 1

2

36

3637

03637

01332

=

=−=

=−+

=−+

=

n

nn

nn

n

6. If 1 + 2 + 3 + ....+ k = 325 , find 13+23 +33 +......+ k3

105625 =

325 =

2

1)+k(k = k +......+ 3 + 2+ 1

2

1)+k(k

325 =k +........... + 3 + 2 + 1

2

23333

= 325

7. If 13 + 23 + 33 +.....+ k3 = 44100.Find 1+2+3 +......+k


210 =k + ......... + 3 + 2 + 1

44100 = 2

1)+k(k

44100= 2

1)+k(k

44100= k +............+ 3 +2 + 1

2

33 33

8. How many terms of the series 13 + 23 + 33 +......... should be taken to get the sum 14400?

15. :terms of number Rqd

15 16,- =n

0 = 15) -16)(n +(n

0 = 240 -n + n2

240 =n + n

2120 = 1) +n(n

14400 = 2

1)+n(n

14400 = 2

1)+n(n

14400 =Sn

2

2

×

.

9. Sum of squares of first n natural numbers is 285, Sum of their cubes is 2025. Find the value of n.

9 =n 90

285 = 1 +2n

6

1) +(2n 906

1) +1)(2n +n(n

285 = n + ............+ 3 + 2 + 1

901)+n(n2

1)+n(n

2025 = 2

1)+n(n

2025 = n + ............+ 3 +2 + 1

2222

2

33 33

6

285

285

45

×

=

=

=

=

10. Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm,…, 24 cm. How much area can be decorated with these colour papers?

2

22222222

2222

cm 4615=

285 - 4900= 6

9(10)(19) -

6

24(25)(49) =

6

1) + 1)(2(9) + 9(9 -

6

1 + 1)(2(24) + 24(24 =

)9 +..... + 3 + 2 + (1 - )24 +..... + 3 + 2 + (1 =

24 +............. + 12 + 11 + 10

area Required

11. Find sum of series (23 − 1) + (43 −33)+(63 −53)+.. to

(i) n terms (ii) 8 terms First numbers are even ; second numbers are odd.

2240 =

192 + 2048 =

3(8) + 4(8)=S

3n + 4ntermsn of Sum

3n + 4n=

n +3n - 3n -2n + 6n + 4n=

n +3n - 3n - 1) +2n +n + 2n(2n =

n + 1) +3n(n - 1) +1)(2n +2n(n =

n + 2

1) +n(n 6 -

6

1) +1)(2n +n(n 12 =

1 +6n - 12n =

1 +6n - 12n + 8n - 8n =

1] -6n + 12n - [8n - 8n =

]1 - 3(2n)(1) + (1) 3(2n) - [(2n) - 8n =

1) -(2n - (2n) = term General

238

23

23

223

22

2

233

233

3233

33

=

××

14.System of Linear Equations in Three

Variables

1. Solve the following system of linear equations

in three variables

a) 6532223 =++=−+=+− zyxzyxzyx ;;

3 = z ,2 = y1, = x

3 = z , 6 z 21

(3),in 2 = y ,1 = xng Substituti

2 = y7 y+ 5

(4),in 1 = xng Substituti

4y20x5-(4)4

(5)-------1 y3x)()(

(3)-------

(2)-------

(4)-------7 y5x)()(

(2)-------

(1)-------

-(3)-

-(2)-

-(1),-

∴

⇒=++

⇒=

=⇒=

=+

=+×

=++

=++

=−+

=++

=−+

=+−

=++

=−+

=+−

11717

1143

28

1432

6

532

21

532

223

6

532

223

xx

yx

zyx

zyx

zyx

zyx

zyx

zyx

zyx

b) 1632925 =+−=+−=++ zyxzyxzyx ;;

(3)-------

(2)-------

(1)-------

1632

92

5

=+−

=+−

=++

zyx

zyx

zyx

1. cartesian product × = × i) a b {1,3,5} {2,3} {(1,2), (1 ... · winglish coaching centre,...

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