arithmetic progression.ppt

8/16/2019 Arithmetic Progression.ppt

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In all the lists above, we see that the successive

terms are obtained by adding a fxed number tothe preceding terms. Such lists are calledARITH!TI" #R$%R!SSI$&S 'or( A#.

So, An Arithmetic #rogression is a list o* numbers

in which each term is obtained by adding a fxednumber preceding term except the frst term.

This fxed number is called the commondi+erence o* the A#.

Remember that it can be positive'(, negative'-(or ero'/(


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0et us denote the frst term o* an Arithmetic#rogression by 'a1(second term by 'a2 (, nthterm by 'ax ( and the common di+erence by d .

The general *orm o* an Arithmetic #rogressionis 3 a , a d , a 2d , a 4d 555555, a 'n-1(d

&ow, let us consider the situation again inwhich ohit applied *or a 6ob and beenselected.

He has been o+ered a starting monthly salaryo* Rs7///, with an annual increment o* Rs8//.what would be his salary *or the f*th year9


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The nth term an o* the Arithmetic #rogression

with frst term a and common di+erence d isgiven by an:a'n-1( d.

n is also called the general term o* the A#. I* thereare m terms in the Arithmetic #rogression , then

am represents the last term which is sometimesalso denoted by l.

The sum o* the frst n terms o* an Arithmetic#rogression is given by s:n;2


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0et us consider an A.#. with frst term ?a@

and common di+erence ?d@ . The frst term : a1 :a / d : a '1-1(d

The second term : a2 : a d : a '2-1(d

The third term : a4 : a 2d : a '4-1(d

The *ourth term : aB :a 4d : a 'B-1(d The nth term : an : a 'n-1(d


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2, C, 1/, 1B5 'i( Here , frst term a : 2, fnd di+erences in

the next terms a2-a1 : C D 2 : B

a4-a2 : 1/ DC : B aB-a4 : 1B D 1/ : B Since the di+erences are common. Hence

the given terms are in A.#.

check whether the term is in A


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&ow let@s try a simple problem3 #roblem 3Eind1/th term o* A.#. 12, 17, 2B, 4/5

Eind the sum o* 4/ terms o* given A.#. 12 2/ 27 4C555

A ladder has rungs 28 cm apart. The rungs decreaseuni*ormly in length *rom B8 cm at the bottom to 28cm at the top .I* the top and the bottom rungs aretwo and a hal* meter apart, what is the length o* the

wood reFuired *or the rungs9

Problems


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%iven A.#. is 12, 17, 2B, 4/.. Eirst term is a : 12"ommon di+erence is d : 17- 12 : C nth term is an: a 'n-1(d

#ut n : 1/, a1/ : 12 '1/-1(C : 12 G x C : 12

8B a1/ : CC Problem 2 Solution 3 %iven A.#. is 12 , 2/, 27 ,

4C Its frst term is a : 12 "ommon di+erence is d :2/ D 12 : 7

The sum to n terms o* an arithmetic progression Sn: n < 2a 'n - 1(d = : x 4/ < 2x 12 '4/-1(x7= : 18 < 2B 2G x7= : 18


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#roblem -4 0et the frst term, a : / and d: common di+erence

:28 cm:/.28m. >hen the rungs are measured*rom top to bottom , then the /, /.28, /.8/, /.8,

1./. 1.28, 1.8/, 1.8, 2.//, 2.28 and 2.8/ meters. JThe number o* rungs is eleven. Rung length

decreases *rom B8 at the bottom to 28 at the top,so rung lengths are3 B8, B4, B1, 4G, 4, 48, 44, 41,2G, 2 and 28 cm.

I* the top and the bottom rungs are two and a hal*meter apart, then The length o* the wood reFuired*or the rungs: 11 x 48 : 478 cm.

Solution

arithmetic progression.ppt

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