Question

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

Answer 1

By using

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ we have,

$S_7=49=\frac{7}{2}(2a+6d)$

$\Rightarrow 14a+42d=98$

$\Rightarrow a+3d=7$ ............. $\left(4^{th}terms\right)$ ........... (i)

$S_{17}=289=\frac{17}{2}(2a+16d)$

$\Rightarrow 34a+272d=578$

$\Rightarrow a+8d=17............\left(9^{th}term\right)$

Deducting $4^{th}$ term from $9^{th}$ term we get

$5d=10$

$\Rightarrow d=2$

Putting the value of d in equation (i) we get $a=1$

Now, $S_n=\frac{n}{2}(2+2(n-1\left)\right)=n(1+n-1)=n^2$