Question

If the sum of the first $7$ terms of an A.P. is $49$ and that of $17$ terms is $289$; find the sum of the first $n$ terms

Answer 1

Let $a$ be the first term and $d$ be the common difference of the A.P.

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

Given that,

$S_7=49$ and $S_{17}=289$

$S_7=7a+21d=49$

$a+3d=7$ $..........\left(1\right)$

$S_{17}=17a+136d=289$

$a+8d=17$ $..........\left(2\right)$

On subtracting eqn.$\left(1\right)$ from eqn.$\left(2\right)$, we get

$5d=10$

So, $d=2$

And putting value of $d=2$ in eqn.$\left(1\right)$, we get

$a+6d=7$

$a=7-6$

$a=1$

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

$=\frac{n}{2}[2+(n-1\left)2\right]$

$=n[1+(n-1\left)\right]$

$=n\times n$

$=n^2$