Question

If the sum of n terms of an A.P. is cn (n - 1), where $c\ne0,$ then the sum of the squares of these terms is

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

Answer 1

Answer: (B)

If $t_{r}$ be the rth 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,
$t^{2}_{1}+t^{2}_{2}+...+t^{2}_{n} =4c^{2}(0^{2}+1^{2}+2^{2}+...+(n-1)^{2})$
$=4c^{2} \frac{(n-1)n(2n-1)}{6}$
$\frac{2}{3} c^{2} n(n-1) (2n-1)$