Question

The arithmetic progression with integers terms, with the property that n is the positive integer n then the sum of first n terms is a perfect square.If true enter 1 else 0.

Answer 1

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

For $n=1S_n$ is a perfect square i.e. $a=k^2$

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

If $S_n$ is to be a perfect square

then $2k^2-d=0(d=2k^2)$

Hence AP become $k^2,3k^2,5k^2,.........(2n-1)k^2$

and $S_n=n^2{k}^2=(nk{)}^2$

where $k$ is a natural number.