I need help with this question involving an A.P.

The question says: If the sum of $\displaystyle n$ terms of an A.P is $\displaystyle cn^2$ then the sum of squares of these $\displaystyle n$ terms will be equal to:

(a)$\displaystyle \frac{n(4n^2-1)c^2}{6}$ (b)$\displaystyle \frac{n(4n^2+1)c^2}{3}$ (c)$\displaystyle \frac{n(4n^2-1)c^2}{3}$ (d)$\displaystyle \frac{n(4n^2+1)c^2}{6}$

What I did: Let the sum of n terms of the A.P be $\displaystyle S_{n}$.

$\displaystyle \Rightarrow S_{n}=\frac{n}{2}[a_{1}+a_{n}]$

$\displaystyle \Rightarrow cn^2=\frac{n}{2}[a_{1}+a_{n}]$

$\displaystyle \Rightarrow 2cn=a_{1}+a_{n}$ ...(i)

When n=1

$\displaystyle \Rightarrow 2c=a_{1}+a_{1}$

$\displaystyle \Rightarrow c=a_{1} ...(ii)$

In equation (i)

$\displaystyle \Rightarrow 2cn-c=a_{n}$ Using (ii)

I have found the value of $\displaystyle a_{n}$. What should I do next?