Question

In the sum fo first n terms of an A.P. 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 that for an $A.P,S_n=c{n}^2$
Then $T_n=S_n-S_{n-1}=c{n}^2-c(n-1{)}^2=(2n-1)c$
$\therefore$ sum of squares of n terms of this A.P.
$=\sum T_n^2=\sum (2n-1{)}^2.c^2=c^2[4\sum n^2-4\sum n+n]=c^2[\frac{4n(n+1)(2n+1)}{6}-\frac{4n(n+1)}{2}+n]=c^{2}n\left[\frac{2(2n^2+3n+1)-6(n+1)+3}{3}\right]=c^{2}n\left[\frac{4n^2-1}{3}\right]=\frac{n(4n^2-1)c^2}{3}$