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Q)

Sum of first $n$ terms of an $AP$ is $kn^2$, sum of their squares will be

$(a)\;\frac{1}{6}nk^2(4n^2+1)\qquad(b)\;\frac{1}{3}nk^2(4n^2-1)\qquad(c)\;\frac{1}{3}nk^2(4n^2+1)\qquad(d)\;\frac{1}{6}nk^2(4n^2-1)$

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A)
Answer : (B) $\;\frac{1}{3}nk^2(4n^2-1)$
Explanation : $a_{n}=S_{n}-S_{n-1}$
$=kn^2-k(n-1)^2$
$=k(2n-1)$
$a_{1}^{2}+a_{2}^{2}+\;...\;+a_{n}^{2}=\sum\;a_{r}^{2}=k^2\;\displaystyle\sum_{r=1}^{n}\;(2r-1)^2$
$=k^2\;[\displaystyle\sum_{r=1}^{n}\;(4r^2-4r+1)]$
$=k^2\;[\frac{4}{6}n(n+1)(2n+1)-\large\frac{4n(n+1)}{2}+n]$
$=\frac{1}{3}k^2n(4n^2-1)\;.$
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