e:iliyas mirza (mathematics wo · 4 answer key & solution 41. fruits are formed from ovary. 42....

2 Answer key & Solution

06. f1 = f

2

m1a

1 = m

2a

2

(m)x = (3m) (y)

x 3

y 1

07. use cosine rule

2 2 2A B Ccos

2AB

2 2 2R R xcos60

2RR

2 2

2

1 2R x

2 2R

x2 = R2

x R

x = 20m

08. Work = f.ds

= N × m

2

mkg m

s

2

2

kg m

s

energy 21

mv2

2m

kgs

2

2

mkg

s

09. At highest point velocity becomes zero but accel-

eration will be non zero

10. L = mv

1 1 1

2 2 2

L m v

L m v

1

2

L 2 4

L 3 5

1

2

L 8

L 15

11. COLM

m1v

1 = (m

1 + m

2)V1

3V4 4 5 6

4

V 20m / s

12. Resistivity does not depend on lenght and cross

sectional area it only depends on material so that

�1 : �

2 :: 1 : 1

13. Bonus

14.

V V1

M + mm

mv (M m)v[COLM]� �

50 × V = (550) × 4

1V 44m / s

mM

V1 V

0 = MV1 + mv

2000V1 = –50×44

1V 1.1m / s recoil velocity of gun

15. In following circuit there is no current is flowing

from the voltmeter so the reading of ideal voltmeter

is 0 (zero)

16. V = IR

log V = log IR

log V = log I + log R

y = mx + C

So graph of log V

and log I is Straight line

17.

r

r

r

n

rR

n

r nR

nR nR nRn

Req = n (nR)

= n2R


18. Memory Based

19. Strong acid + Weak base � Acidic salt

pH is less than 7.

Acidic solution

20.1

Basic StrengthOxidation Number

21. 1

2 3 4

55 4

2 1

3,3-diethyl pentane

22. KetoneC

O

23. Base which is dissolved in water is known as

Alkali.

Fe(OH)2 but does not dissolve in water.

24.

CaCO3 CaO + CO2heat

(Thermal decomposition) due to heat

25. LiAlH4, NaBH

4, Na/C

2H

5 – All

26. Allotropes of Carbon - Diamond, Graphite, Fullerene

27. 0.001M HCl

[H+] = 0.0001 M

[H+] = 1 × 10–4 M

pH = –log [H+]

= – log[1×10–4]

= (–(–4)) log10

= 4 × 1 = 4

28. Na3PO

3 does not contain any hydrogen to donate.

So it is not an acid.

29. When heat is produced along with the product then

it is known as Exothermic reaction.

30. pH = 4 = 10–4 M

pH = 2 = 10–2 M

= 10–2 – 10–4

= 10[–2–(–4)]

= 102

31. CaO + H O2 Ca(OH) + heat2

Change in temperatureof mixture

32. Respiration food burn in presenceof oxygen

release heat

exothermic

33. In oxidation, Metal release electron.

Na – 1e– ��Na+

Positive charge increase

Mg+ – 1e– � Mg+2

Addition of oxygen 2Ca + O2 � 2CaO

34. Acid – H+ = Conjugate base

2– 3–4 4HPO H PO

35. During ovulation, the mature follicle or graafian

follicle burst and the ovum is released. After

ovulation, the grarulosa cells and the interstitial cells

form a mass of large and yellowish conical cells.

This is named as corpus luteum which is serve as

temporary endocrine gland by releasing

progesterone and extrogen.

36. Menstrual cycle - It is the regular natural change

that occurs in the female reproductive system

(specifically-uterus and ovaries) that makes

pregnancy possible. The cycle is required for the

production of oocytes and for the preparation of

uterus for pregnancy.

It is divided inti following phase.

(i) Follicular or proliferative phase

(ii) Luteal or secretory phase

(iii) Secretory phase is also known as luteat phase

and lasts for 13-14 days.

37. Spermiogenesis - It is the final differentiation and

maturation process of the spermatids into sperm

cells.

38. A, B, AB and O

39. Amniocentesis is a medical procedure used

primarily in prenatal diagnosis of chromosomal

abrormalities and faetal infection oas well as for

sex deternimation. Jaundice is not detected by

amniocentesis.

40. Down's syndrome - Genetic disorder caused when

abnormal cell division results in extra genetic

material from chromosome 21.


41. Fruits are formed from ovary.

42. IUCD - Also known as intrauterine contraceptive

device is a small often T-shaped birth control divice

that is inserted into a women's uterus to prevent

pregnancy.

43. Zygote, Morula, Blastula, Implantation

44. Antidiuretic hormone.

45. Melatonin - It is secreted by pineal gland. It regu-

late the normal rhythm of sleepwake cycle.

46. Sympathetic nervous system is part of the ANS.

The sympathetic nervous system activates what is

often termed flight or fight response.

47. Those organism which excrete uric acid is called

uricotelic organisms.

Ex. - Birs and Reptiles

48. Ketones

49. Glucagon is secreted by � cells of pancrease.

50. Memory based

51. 101–1 = 9

102–1 = 99

10n–1 = 999 .... n times

So sum of the digits of 10n–1 = 9+9+9...... n times

� 9n = 3375

3375n 375

9

52. 3sin� + 5cos� = 5 .... (1)

square both sides

� (3sin� + 5cos�)2 = 52

� 9sin2� + 25cos2� + 30 sin�cos� = 25 .... (2)

(5sin� – 3cos�)2 = 25sin2� + 9cos2� – 30sin�cos�from eq. (2) : 25sin2�+9cos2�+9 sin2�+25cos2�–25

(5sin� – 3cos�)2 = 9

� 5sin� – 3cos� = �3

53. 7x – 16y = 0 .... (1)

� 7x = 16y

� 7x = 16 x4

49

� 7x.49 = 16.4x

� 7x+2 = 4x+2

x + 2 = 0

x = –2

4x – 49y = 0 .... (2)

4–2 – 49y = 0

� 1

49y 016

� 1 1

y16 49 784

so (y–x) = 1 1 1569

784 2 784

54. In �ACB :E

30º

60º

60º 30ºBA 20

C

D

F

60º

ACsin30º

AB

� 1 AC

2 20

AC = 10 units

So AD = DC = 5 units

so quadrilateral ACEF will be parallelogram

so AC = EF = 10 units

In �CDE :

DEtan 60º

DC

� DE3

5

� DE 5 3 units

In �DEF : By pythagores theorem� DF2 = EF2 + DE2

� DF2 = 2

210 5 3

DF = 5 7 units so x = 7

55. (1 + 2x)20 = a0 + a

1x + a

2x2 + ....... + a

20x20

put x = 1

320 = a0 + a

1 + a

2 + ...... + a

20 .... (1)

eq. (1) multiply by 5

5.320 = 5a0 + 5a

1 + 5a

2 + ...... + 5a

20 .... (2)

put x = –1

1 = a0 – a

1 + a

2 – a

3 + ...... + a

20 .... (3)

eq. (2) + eq. (3) :

5.320 + 1 = 6a0 + 4a

1 + 6a

2 + 4 a

3 +...... + 6a

20

5.320 + 1 = 2 (3a0 + 2a

1 + 3a

2 + 2a

3 + ...... + 3a

20)


� 3a0 + 2a

4 + 3a

2 + 2a

3 + .... 3a

20 =

205.3 1

2

56. R1

A C

R2B

r

P R Q

��

�R1

A B

r

P PR Q

L

B

r R2

C

M+

AB = R1 + r BC = R

2 + r

AL = R1 – r MC = R

2 – r

PR = LB RQ = BM

In �ALB :

AB2 = LA2 + LB2

� LB2 = AB2 – LA2

� LB2 = (R1 + r)2 – (R

1 – r)2

� LB2 = 4R1r

PR = LB = 12 R r ....(1)

In �CMB :

BM2 = BC2 – MC2

� BM2 = (R2 + r)2 – (R

2 – r)2

� BM2 = 4R2r

RQ = BM = 22 R r ....(1)

R1

AC

R2

P Q

X y

AC = R1 + R

2

AX = R1 – R

2 (PQ = XY)

�AXY :

XY2 = AC2 – AX2

XY2 = (R1+R

2)2 – (R

1–R

2)2

� PQ = XY = 1 22 R R .... (3)

� PQ = PR + RQ

� 1 2 1 22 R R 2 R r 2 R r {dividing by 1 2R R r

�1 2

1 1 1

r R R

57. f(x) = x5 + ax4 + bx3 + cx2 + d

f(1)=1, f(2)=2, f(3)=3, f(4)=4, f(5)=5

Let a polynomial g(x) such that it has 5 roots which

are x = 1, 2, 3, 4, 5

� g(x) = f(x) – x

� (x–1) (x–2) (x–3) (x–4) (x–5) = f(x) – x

� f(x) = (x–1) (x–2) (x–3) (x–4) (x–5) + x

to find d put x = 0

d = (–1) × (–2) × (–3) × (–4) × (–5)

d = –120

58. f(x) is a 3rd degree polynomial

let f(x) = k(x–a) (x–b) (x+c) .... (1)

and a,b,c �R+

if f(x2) = k(x2–a) (x2–b) (x2+c)

2 2 2

2 2 2

x a 0 x b 0 x c 0

x a x b x c

x a x b x c

Two roots Two roots No real roots

so f(x) has all three real roots.

59. Let first term = a,

x

xS 2a x 1 2

2

��x

49 2a x 1 22

��49 = x[a + x – 1]�� x = 7Common difference = d = 2

a7 = a + 6d

� 13 = a + 6×2

a = 1

60. Let first term = a, common difference = d

s2n

= 3sn

� 2a 2n 1 d 3 2a n 1 d

� 4a + (4n–2)d = 6a + (3n–3)d� 2a = (n–1)d .... (1)

3n

n

3 2a 3n 1 dS

S 2a n 1 d

6a 9n 3 d

2a n 1 d put from eq. (1)

3n 3 d 9n 3 d

n 1 d n 1 d = 6

61.2 3 31 2

2 3 31 2

x x x 1x x1Area

y y y 1y y2


2 2 a b b c c 11 1a b

b a a c c 12 2

2 2 2 21 1a b a b bc ac 0

2 2

2 2 2 2a b a b c b a

c = 0

62.

(8,0) BA (0,0)

OC (0, 15)

Y

X

2 2BC 8 0 15 0

BC 64 225

BC = 17 units

In right angled triange :

Circum radius = Hypotenuse

2

� Circum radius =17

8.5 units2

63.C

(x, y)(5, –9)(–2, 5)

A B3 : 4

3 9 4 55 3 4x 2x , y

3 4 3 4

x = 1 y = –1Ans = (x, y) = (1, –1)

64. Let the sides of a triangle are a,b,c.

a = 5cm a+b+c = 16cm so b + c = 11 cm

Case-I :

b + c = 11

a = 5 b = 5 c = 6

5 5 6S 8

2

Area = s s a s b s c

Area = 12 sq. cm

65.

D CE

A 6 cm B

F

Area of square ABCD = 62 = 36 sq.cm

ar (�ABF) = ar (�BCE) = ar (�FBED) = 12 sq.cm

ar BDE DE

ar BCE EC�

��6 DE DE 1

12 EC EC 2

So CE = 4cm

In �BCE = BE2 = CE2 + BC2

� BE2 = 42 + 62

BE 2 13 cm

66. Area of shaded region =

Area of quadrant – Area of �ABC

21r 3 4

4

22512 cm

4

67. tan45º = tan(18º +27º)

tan18º tan 27º1

1 tan18º tan 27º

tan A tan Btan A B

1 tan A tan B�

� 1 – tan18ºtan27º = tan18º + tan27º� tan18º + tan27º + tan18ºtan27º = 1

68. tan(�cos�) = cot (�sin�)

� tan(�cos�) = tan sin2

� �cos� = sin2

� 1

sin cos2

��

from eq. (1)

1 1 1sin cos .

22 2

1 1 1sin cos

2 2 2 2

1sin cos cos sin

4 4 2 2

1sin

4 2 2


69. 1 + sec� + tan� = p

� sec� + tan� = P–1 ....(1)

� sec� – tan� = 1

P 1....(2)

sec2� – tan2� = 1

� (sec��tan�) (sec�–tan�) = 1

� (sec�–tan�) = 1

sec tan

eq. (1) + eq. (2)

12sec P 1

P 1

2P 2P 22sec

P 1

2

2 P 1cos

P 2P 2

So sin� = 21 cos

� sin� =

2

2

2 P 11

P 2P 2

� sin� = 2

2

P 2P

P 2P 2

70.

B D C

45º

45º 45º

A

E

From given figure :

AE = DE ... (1)

A E

B DC Cº º

�ABC and �DEC :

�ABC � �DEC (By AAA Rule)

AB AC

DE DE

AB DE AC CE

AC CE AB DE

AC CE1 1

AB DE

AB AC DE CE

AB DE

AB AC AC

AB DE

DE AB AC AC AB

71.

A

r

r

Orp

B D1

C

E30

º

�ADOE will be square

so BD = (r–1) units

In �ABC : AC

3AB

AC 3AB

In �POB :

PBtan 30º

OP

1 r 1

r3

1 11

r3

1 3 1

r 3

3 3 1r

3 1 3 13 3

r units2

72.60 12

40 x

� x = 8 cm

73. (x–1) (x–2) (3x–2) (3x–1) = 12

� (x–1) (3x–2) (x–2) (3x+1) = 12

�� (3x2–5x+2) (3x2–5x–2) = 12

Let 3x2 – 5x = A

�� (A+2) (A–2) = 12

��A2–4 = 12

��A= � 4

A = 4

��3x2 – 5x = 4


��3x2 – 5x – 4 = 0

D = b2 – 4ac

D = 25 + 48 = 73

D > 0

two irrational roots

A = –4

��3x2 – 5x = –4

��3x2 – 5x + 4 = 0

D = b2 – 4ac

D = 25 – 48

D < 0

No real roots

74.1 1 1

S .......1.6 6.11 11.16

1 1 1 11 1 1S 1 .....

5 6 6 11 11 16

1S

5

75. y y y........... t

�� yt t

�� 2yt t

�� t = y

x x x....... u

�� xu u

�� 2xu u�� u = x

xy

y y y......

log xlog x x x....... log

log y

76. L is a mid point of PQ

� ar (�PQR) = 108 cm2

So ar (�LQR) = ar (�PLR) = 54 cm2

� LK || MR

So ar (�LKR) = ar (�LMK) .... (1) (Same base

and between parallel lines)

� ar (�LQR) = ar (QLK) + ar (�LKR)

= ar (�QLK) + ar (�LMK)

ar (�LQR) = ar (�QMK) = 54 cm2

77. x4 – y4 = 25 × 7

� (x2–y2) (x2+y2) = 25 × 7

� (x + y) (x – y) (x2 +y2) = 25 × 7

� (x – y) (x + y) (x2 + y2) = 1 × 7 × 25

So x – y = 1 .... (1)

x + y = 7, x2 + y2 = 25 .... (2)

x = y + 1 put into eq. (2)

y = –4 y = 3 x = � 3 x = � 4

(3, –4) (–3, –4) (4, –3) (–4, 3)

78. f(x) = x3 + bx2 + cx + d

f(1) = 0 f(1) = 0f(0) = 2

b + c + d = 1 ..... (1) 1 + b + c + d = 0

b – c + d –1 = 0 ..... (3) –1 + b – c + d = 0

d = 2 ..... (2)

from (1), (2), (3) we get b = –2

79. y = x2 + px + 12 .... (1) it has a root = 4

� (4)2 + P(4) + 12 = 0

� P = –7

y = x2 + px + q .... (2) it has equal zeros.

� y = x2 –7x + q

So D = 0

� b2 – 4ac = 0

� 49 – 4q = 0 � 49

q4

80. B

(–4,0) (6,0)

C

(2,3)A

(0,2)

180,

4

3x + 4y – 18 = 0 ...... (1)

x – 2y + 4 = 0 .... (2)

from eq. (1) and eq. (2) :

x = 2 y = 3

so point A (2, 3)

Area of �ABC = 1

2

2 3 1

4 0 1

6 0 1

= 15 sq. units

81. 2x + (k2 + 1)y – 3 = 0 ..... (1)

x + ky – 4 = 0 ..... (2)

condition for no. solution : 1 1 1

2 2 2

a b c

a b c

22 k 1

1 k

� k2 + 1 = 2k � k2 –2k + 1 = 0k = 1


82. Selecting 5 number as table :

Total no. of ways (�) = 196 + 192 + 188 +.....+ 4

49

2(196+4) = 4900 ways

83.20

20

1 3 7 15 2 1S ........

2 4 8 16 2

19 20

20 21

1 1 3 7 2 1 2 1S ........

2 4 8 16 2 2

=20

21

1 1 1 1 1 2 1S ........ 20 times

2 2 2 2 2 2

20

21

1 20 2 1S

2 2 2

20

20

2 1S 20

2

20 20

20

20.2 2 1S

2

2020

20

19.2 1S 2 19

2

so � = 20

84. rr 1

rS

2 1

2 3 4

1 2 3 4 5S .......

2º 2 2 2 2

2 3 4

1 1 2 3 4S ............

2 2 2 2 2

Such that 2 3

1 1 1 1S 1 ............

2 2 2 2

S = 4

85. sin2x – 3sinx + 2 = 0

� sin2x – 2sinx – sinx + 2 = 0

� sinx (sinx–2) – 1 (sinx – 2) = 0

sinx = 1 or sinx = 2

So x2

17 sin3x + 2 = 317sin 2

2 = 19

86.

RQ

P

r 2

r 1

D2

D1

QS = 15 units

Let the inradius of �PQR = r

the inradius of �PQS = r1

the inradius of �QSR = r2

r + r1 = 10 ...... (1)

1r PQ QR PR

2 ..... (2)

1

1r PS QS PQ

2 ..... (3)

from (1), (2), (3) :

PQ + QR – PR + PS + QS – PQ = 20

� PS + QR + QS – PR = 20

� PS + QR – PR = 5

� QR – SR = 5

So 2

1r QS RS QR

2

2

1r 15 RS QR

2

r2 = 5 units

So r2 × 4 = 5 × 4 = 20

87.

D

C

B

A

F

E

m

1

2

23

3

1

1 1

2

23

3

�1 + �

2 + �

3 = 180º

1 1sin

2 2m ..... (1)


2 1sin

2 m ..... (2)

3 3sin

2 2m ..... (3)

1 2180º 3sin

2 2m

1 2 3cos

2 2m

1 2 1 2 3cos cos sin sin

2 2 2 2 2m

2 2

1 1 1 1 31 1

2m m 2m4m m

2 2

2

4m 1 m 1 3 1

2m m 2m 2m

2 2

2 2

m 1 4m 1 3m 1

2m 2m

4 24m 14m 6m 032m 7m 3

10m3 – 35m = 15

88.

B C

A D

10 zz18

0º-c

5

5C

M

C

Ey

y

In �ABC : AB = AC = 10 cm

BC = 6 cm

�B = �C

by cosine Rule :

2 2 210 6 10cosc

2 6 10

3cosc

10

In �ACE :

22 225

5 z3

cos 180º c25

2 53

262525 z3 9

2510 2 53

2 1075z

9

In �MBE :

2 225 y ycosc

2 5 y

3 25

10 10xy

25y

3

In �ADC :

2 1075100 AD

9cosc2 10 AD

29AD 54AD 175 0

25AD

3So 3(AD)

253 25

3

89. Let radius of bigger circle is = R units

Area of bigger circle = 8�

� �R2 = 8�� R 2 2 units

from given figure :

r2 + r2 = (R–r)2

� 2r2 = R2 + r2 – 2Rr

� r2 = R2 – 2Rr

� 2r2 = 8 – 4 2 r

� 2r 4 2r 8 0

r 2 2 4

So 2 2 r 2 2 2 2 4 4


90. A + B + C = 180º .... (1)

cos A cos B cosC

sin Bsin C sin Csin A sin Asin B

cos 180º B C cos 180º A C

sin BsinC sin Asin C

cos 180º A B

sin Asin B

cos B C cos A C cos A B

sin Bsin C sin Asin C sin Asin B

cos BcosC sin Bsin C cosAcosC sin Asin C

sin Bsin C sin Asin C

cos Acos B sin Asin B

sin Asin Bsin A cosBcosC sin Asin Bsin C cosA sinBcosC sin Asin Bsin C cosA cosBsin C sinA sinBsinC

sin Asin Bsin C

3sin Asin Bsin C sin A cos BcosC cos Asin BcosC cos AcosBsin C

sin Asin Bsin C

sin(A+B+C) = sinAcosBcosC + cosAsinBcosC +cosAcosBsinC-sinAsinBsinCso sinAcosBcosC + cosAsinBcosC+cosAcosBsinC= sinAsinBsinC

= 2sin Asin Bsin C

2sin Asin Bsin C

91. tan3A = 4tanA

sin3A sin A4

cos3A cos A

sin 3A cos3A4

sin A cos A

2cosA 4cos A 3sin 3A4

sin A cos A

2sin3A4 4cos A 3

sin A .... (1)

sin3A 8

sin A 3

30sin 3A 830 80

sin A 3

3cos3A 4cos A 3cosA

tan 3A 4 tan A

3

2

3tan A tan A4tan A

1 3tan A

2 23 tan A 4 12 tan A

211tan A 1

1tan A

11

2 11cos A

12�

92. let g(x) = f(x) + x2 ..... (1)

so g(x) has four roots are –1,2, –3 and 4.

2x 1 x 2 x 3 x 4 f x x

2f x x 1 x 2 x 3 x 4 x ....(2)

Put x=1 f 1 2 1 4 3 1

93. The fraction 2 29a 14b

9abmust be an intger

So 2 2 29 | 9 14 9 | 3 |a b b b

since b=3n

Put b=3n given expression we get 2 2a 14n

3an

So 2 2 2n | a 14n n | a

but gcd(a,b)=1 � gcd(a,3n)=1 � gcd(a,n)=1

so n=1 and b=3

put b=3 we get 2a 14

3a, so a is a factor of 14.

a = {1,2,7,14}

So four solutions are (1,3), (2,3), (7,3), (14,3)

94. n nna 6 8

83 8383a 6 8

83 8383a 7 1 7 1

n n n 0 n n 1 n n 2 2 n 0 n0 1 2 na b c a b c a b c a b ..... c a b

83 82 80 183 83 83 8383 0 1 2 1a c 7 c 7 c 7 ......... c 7 1

83 82 81 80 183 83 83 83 830 1 2 3 4c 7 c 7 c 7 c 7 ..... c 7 1

83 81 283 83 83 8383 0 2 2 1a 2. c 7 2. c 7 ....2. c 7 2. c 7

so 2 283 8349 | a 7 | a 7 | 2.83.7

we get remainder=35

95. Let x = 222 .... 300 times

by Modulo Rule :

x � 6 (Mod 9) ... (1)

x � 0 (Mod 111) ... (2)

from eq. (2) x = 111a put into eq. (1)

� 111a = 6 (mod 9)

a = 9b + 2


So x = 111 (9b+2)

x = 999b + 222 so remainder =222

96. LCM (a,b) = 23×53

LCM (b,c) = 24×53

LCM (c,a) = 24×53

Calculation of power of 2 :

Case-I : c=4, a=3, b=0,1,2,3

Total options is = 4

Case-II : c=4, b=3, a=0,1,2

Total options are = 3

So total cases of power of 2 = 4 + 3 = 7

Calculation of power of 5 :

Case-I : a=4, b=3, c=0,1,2,3


Case-II : a=0,1,2 b=3, c=3


Case-III : a=3, b=0,1,2 c=3

So total cases of power of 5 = 10

� Total combinations are 7 × 10 = 70

97.n

2r

r 1

1t n 2n 9n 13

6

Let tr = ar2 + br + c

r = 1 a+b+c=4 .... (1)

r = 2 4a+2b+c=9 .... (2)

r = 3 9a+3b+c=16 .... (3)

To solve eq. (1), (2) and (3)

we get a=1, b=2, c=1

so tr = r2 + 2r + 1

tr = (r+1)2

r 1 r 1r

1 1

r r 1r. t

r 1

1 1

r r 1

1 1 1 1 11 ...

2 2 3 3 4

= 1

98. Let x = 48199

x � 0 (mod 4) .... (1)

x � 12 (mod 25) .... (2)

from eq. (2) x = 25a + 12 .... (3)

from eq. (1) 25a + 12 = 0 (mod 4)

a = 4b

from eq. (3) x = 25(4b) + 12

x = 100b +12

So remainder = 12

99. sin� + cos� = p .... (1)

2 2sin cos p

21 2sin cos p

22p1 p

q

sec� + cosec� = q

1 1q

cos sin

sin cosq

sin cos

psin cos

q .... (2)

2 22pp 1 q p 1 2p

q

100.n k

2 4 3 2

k 1 m 1

m an bn cn dn e

L.H.S. = n

k 1

k k 1 2k 1

6

n4 3 2 3 2

k 1

1an bn cn dn e 2k 3k k

6

22n n 1 n n 1 2n 1 n n 112. 3.

6 2 6 2

221n n 1 n n 1 2n 1 n n 1

6

24 3 2 21an bn cn dn e n n 1 n n 1 2n 1 n n 1

12

put n =1 a+b+c+d+e =1

4 6 2 112

e:iliyas mirza (mathematics wo · 4 answer key & solution 41. fruits are formed from ovary. 42....

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