Question

If the sum of n terms of an A.P. is $cn(n-1)$, where $c\ne 0$, then the sum of the squares of these terms is?

• A. $c^{2}n(n+1{)}^2$
• B. $\frac{2}{3}{c}^{2}n(n-1)(2n-1)$
• C. $\frac{2c^2}{3}n(n+1)(2n+1)$
• D. None of these

Answer 1

Answer: (B)

$\frac{2}{3}{c}^{2}n(n-1)(2n-1)$

If $t_r$ be the $r^{th}$ term of the A.P., then
$t_r=S_r-S_{r-1}$
$=cr(r-1)-c(r-1)(r-2)$
$=c(r-1)(r-r+2)=2c(r-1)$
We have,
$t_1^2+t_2^2+\cdots +t_n^2=4c^2(0^2+1^2+2^2+\cdots +(n-1{)}^2)$
$=4c^2\frac{(n-1)n(2n-1)}{6}$
$=\frac{2}{3}{c}^{2}n(n-1)(2n-1)$