IIT-JEE 2009 · Paper 2 · Mathematics Q20

MathematicsSingle correct+3 / -1Question 20 of 57

If the sum of first $n$ terms of an A.P. is $cn^2$ , then the sum of squares of these $n$ terms is

\dfrac{n(4n^2-1)c^2}6

\dfrac{n(4n^2+1)c^2}3

\dfrac{n(4n^2-1)c^2}3

\dfrac{n(4n^2+1)c^2}6

Answer and worked solution

Official answer: C

Terms. $a_k=S_k-S_{k-1}=c\left[k^2-(k-1)^2\right]=c(2k-1)$ .

Sum of squares.
$\sum_{k=1}^n c^2(2k-1)^2=c^2\cdot\frac{n(2n-1)(2n+1)}3=\frac{n(4n^2-1)c^2}3.$

(C)

Topics:Special series sums Arithmetic progression

IIT-JEE 2009 · Paper 2, Mathematics question 20