Question

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

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

Answer 1

Answer: (B)

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

Let the $A.P$ be $a,a+d,a+2d,\dots ,a+(n-1)d$

Given $S_n=cn(n-1)$

$cn(n-1)=\frac{n}{2}(2a+(h-1\left)d\right)$

$2c(n-1)=2a+(n-1)d$

$2c=d$

$-2c=2a-d$

$-2c=2a-2c$

$a=0$

Required series

$o,d^2,(2d{)}^2,........((n-1)d{)}^2$

Sum of $(n-1)$ terms square

$=o+d^2+(2d{)}^2+........+((n-1)d{)}^2$

$=d^2[1^2+2^2+3^2+\cdots +(n-1{)}^2]$

$=\sum \left[\right(n-1)\times d{]}^2$

$=\sum \left[\right(n-1)\times 2c{]}^2$

$=4c^2\times \sum (n-1{)}^2$

$=\frac{4c^2\times (n-1)n(2n-1)}{6}$ since sum of square of first n natural numbers $S=\frac{n(n-1)(2n-1)}{6}$