Question

Any odd square i.e ${[2n+1]}^2$ is equal to sum of n terms of an A.P. in increased by unity. Find the first term of that A.P ?

Answer 1

Let $a$ be the first term and $d$ be the common difference of the A.P
Given $(2n+1{)}^2=\frac{n}{2}[2a+(n-1)d]+1$
$\Rightarrow 4n^2+4n+1=\frac{d}{2}{n}^2+\frac{2a-d}{2}n+1$
Comparing coefficient of $n$ and $n^2$ we get $d=8,\frac{2a-d}{2}=4\Rightarrow a=8$
Hence first term of the A.P is 8.