Question

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

Answer 1

Given that, $S_7=49and{S}_{17}=289$
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\therefore S_7=49=\frac{7}{2}[2a+(7-1\left)d\right]\Rightarrow 49=\frac{7}{2}=(2a+6d)\Rightarrow (a+3d)=7......\left(i\right)Similarly,S_{17}=\frac{17}{2}[2a+(17-1\left)d\right]\Rightarrow 289=\frac{17}{2}[2a+16d]\Rightarrow (a+8d)=\mathrm{17......}\left(ii\right)$
Substracting equation (i) from equation (ii), we get that 5d = 10.
Therefore, the value of d = 2 and a = 7-3d=1
$\therefore S_n=\frac{n}{2}[2a+(n-1\left)d\right]=\frac{n}{2}\left[2\right(1)+(n-1\left)\right(2\left)\right]$
$=\frac{n}{2}(2+2n-2)=\frac{n}{2}\left(2n\right)=n^2$
Therefore, the sum of n terms of the AP is $n^2$