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Question

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

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Solution

It is given that sum of first 7 terms of an AP is equal to 49 and sum of first 17 terms is equal to 289.

Applying formula, Sn=n2(2a+ (n−1) d) to find sum of n terms of AP , we get

49=72(2a + (7−1) d)

98=7(2a+6d)

98=14a+42d

7=a+3d

a=7−3d (1)

And, 289=172(2a+ (17−1) d)

578=17(2a+16d)

34=2a+16d

17=a+8d

Putting equation (1) in the above equation, we get

17=7−3d+8d

10=5d

d=105=2

Putting value of d in equation (1), we get

a=7−3d=7−3 (2) =7−6=1

Again applying formula, Sn=n2(2a+ (n−1) d) to find sum of n terms of AP , we get

Sn=n2[2(1) + (n−1) (2)]

Sn=n2[2+2n−2]

Sn=n2[2n]

Sn= n2

Therefore, sum of n terms of AP is equal to n2


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