Search any question & find its solution
Question:
Answered & Verified by Expert
If the sum of first $n$ terms of an $\mathrm{AP}$ is $\mathrm{cn}^2$, then the sum of squares of these $n$ terms is
Options:
Solution:
1485 Upvotes
Verified Answer
The correct answer is:
$\frac{n\left(4 n^2-1\right) c^2}{3}$
$\frac{n\left(4 n^2-1\right) c^2}{3}$
Let $S_n=c^2$
$$
\begin{aligned}
& S_{n-1}=c(n-1)^2=c n^2+c-2 c n \\
& \therefore T_n=2 c n-c \quad\left(\because T_n=S_n-S_{n-1}\right) \\
& T_n^2=(2 c n-c)^2=4 c^2 n^2+c^2-4 c^2 n \\
& \therefore \text { Sum }=\Sigma T_n^2=\frac{4 c^2 \cdot n(n+1)(2 n+1)}{6} \\
& =\frac{2 c^2 n(n+1)(2 n+1)+3 n c^2-6 c^2 n(n+1)}{3} \\
& =\frac{n c^2\left[4 n^2+6 n+2+3-6 n-6\right]}{3} \\
& =\frac{n c^2\left(4 n^2-1\right)}{3}
\end{aligned}
$$
$$
\begin{aligned}
& S_{n-1}=c(n-1)^2=c n^2+c-2 c n \\
& \therefore T_n=2 c n-c \quad\left(\because T_n=S_n-S_{n-1}\right) \\
& T_n^2=(2 c n-c)^2=4 c^2 n^2+c^2-4 c^2 n \\
& \therefore \text { Sum }=\Sigma T_n^2=\frac{4 c^2 \cdot n(n+1)(2 n+1)}{6} \\
& =\frac{2 c^2 n(n+1)(2 n+1)+3 n c^2-6 c^2 n(n+1)}{3} \\
& =\frac{n c^2\left[4 n^2+6 n+2+3-6 n-6\right]}{3} \\
& =\frac{n c^2\left(4 n^2-1\right)}{3}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.