The sum  of  n terms  of  an  AP is an(n - 1). The  sum of  the  squares  of these  terms  is

Since,${\mathrm{S}}_{\mathrm{n}}=\mathrm{an}(\mathrm{n}-1)$ 

Now, nth term of the series is

 ${\mathrm{T}}_{\mathrm{n}}={\mathrm{S}}_{\mathrm{n}}-{\mathrm{S}}_{\mathrm{n}-1}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{T}}_{\mathrm{n}}=\mathrm{an}(\mathrm{n}-1)-\mathrm{a}(\mathrm{n}-1)(\mathrm{n}-2)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{T}}_{\mathrm{n}}=\mathrm{a}(\mathrm{n}-1)[\mathrm{n}-(\mathrm{n}-2\left)\right]=2\mathrm{a}(\mathrm{n}-1)\end{array}$

Again, let the sum of squares of n terms of the series is
S₁ such that

$\mathrm{S}={\mathrm{T}}_1^2+{\mathrm{T}}_2^2+{\mathrm{T}}_3^2+\dots +{\mathrm{T}}_{\mathrm{n}}^2=\sum _{\mathrm{r}=1}^{\mathrm{n}} {\mathrm{T}}_{\mathrm{r}}^2$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{S}=\sum _{\mathrm{r}=1}^{\mathrm{n}} \left\{2\mathrm{a}\right(\mathrm{r}-1){\}}^2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{S}=\sum _{\mathrm{r}=1}^{\mathrm{n}} 4{\mathrm{a}}^2(\mathrm{r}-1{)}^2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left.\mathrm{S}=4{\mathrm{a}}^2\left[\frac{1}{6}(\mathrm{n}-1)\mathrm{n}\left\{2\right(\mathrm{n}-1)+1)\right\}\right]\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{S}=\frac{2{\mathrm{a}}^2}{3}\mathrm{n}(\mathrm{n}-1)(2\mathrm{n}-1)\end{array}$