Question

If the sum of the first $n$ terms of an $AP$ is $c{n}^2$ then the sum of squares of these $n$ terms is

• A. $\frac{n(4n^2-1)c^2}{6}$
• B. $\frac{n(4n^2+1)c^2}{3}$
• C. $\frac{n(4n^2-1)c^2}{3}$
• D. $\frac{n(4n^2+1)c^2}{6}$

Answer 1

The correct option is C $\frac{n(4n^2-1)c^2}{3}$

Given
$\frac{n}{2}(2a+(n-1\left)d\right)=c{n}^2$
$\Rightarrow a+a_n=2cn$
$\Rightarrow a_n=2cn-a$____(1)
$\Rightarrow a_n^2=4c^2{n}^2+a^2-4cna$
$\therefore \sum a_n^2=4c^2\sum n^2+\sum a^2-4ca\sum n$
$=4c^2\frac{n(n+1)(2n+1)}{6}+n{a}^2-4ca\left(\frac{n(n+1)}{2}\right)$
$=\frac{2c}{3}\left\{n\right(n+1\left)\right((2n+1)c-3a\left)\right\}+n{a}^2$____(2)
Now, from (1) , $a_n=2cn-a$
$\Rightarrow \sum a_n=2c\sum n-\sum a$
$\Rightarrow c{n}^2=2\frac{cn(n+1)}{2}-an=c{n}^2+cn-an$
$\Rightarrow cn-an=0$ or $c=a$
Substituting in (2),
$\sum a_n^2=\frac{2c}{3}\left\{n\right(n+1\left)\right((2n+1)c-3c\left)\right\}+n{c}^2$
$=\frac{2c^{2}n(n+1)(2n-2)+3n{c}^2}{3}$
$=\frac{n{c}^2(4n^2-4+3)}{3}$
$=\frac{n{c}^2(4n^2-1)}{3}$