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.

Answer 1

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, $S_n$=$\frac{n}{2}$(2a+ (n−1) d) to find sum of n terms of AP , we get

49=$\frac{7}{2}$(2a + (7−1) d)

$\Rightarrow$98=7(2a+6d)

$\Rightarrow$98=14a+42d

$\Rightarrow$7=a+3d

$\Rightarrow$a=7−3d (1)

And, 289=$\frac{17}{2}$(2a+ (17−1) d)

$\Rightarrow$578=17(2a+16d)

$\Rightarrow$34=2a+16d

$\Rightarrow$17=a+8d

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

17=7−3d+8d

$\Rightarrow$10=5d

$\Rightarrow$d=$\frac{10}{5}$=2

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

a=7−3d=7−3 (2) =7−6=1

Again applying formula, $S_n$=$\frac{n}{2}$(2a+ (n−1) d) to find sum of n terms of AP , we get

$S_n$=$\frac{n}{2}$[2(1) + (n−1) (2)]

$\Rightarrow$$S_n$=$\frac{n}{2}$[2+2n−2]

$\Rightarrow$$S_n$=$\frac{n}{2}$[2n]

$\Rightarrow$$S_n$= $n^2$

Therefore, sum of n terms of AP is equal to $n^2$