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If the sum of the first $n$ terms of an arithmetic progression is $cn^2$, then the sum of squares of these $n$ terms is

  1. $\frac{n(4n^2-1)c^2}{6}$
  2. $\frac{n(4n^2+1)c^2}{3}$
  3. $\frac{n(4n^2-1)c^2}{3}$
  4. $\frac{n(4n^2+1)c^2}{6}$
in Numerical Ability
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1 Answer

1 vote

Answer: $\mathbf C$

Explanation:

$\mathrm {S_n = cn^2} \text{[Given]}$

Now $\mathrm {n^{th}}$ term from the sum can be calculated by using, the formula:

$\mathrm {\therefore t_n = S_n - S_{n-1} = cn^2-c(n-1)^2 = c(2n-1)}$

Now,

$\mathrm {\sum t_{n^2} = \sum c^2(2n-1)^2 = \sum c^2(4n^2+1-4n)}\tag{i}$

Now, we know that:

$\mathrm {\sum n^2 = \frac{n(n+1)(2n+1)}{6}\;,\; \sum n = \frac{n(n+1)}{2}\;, and\;\sum 1 = n}$

Substituting these values in $\mathrm {(i)}$, we get:

$\mathrm {\sum t_{n^2} = c^2\Big(4\frac{n(n+1)(2n+1)}{6}\;+n + \frac{n(n+1)}{2}\Big) = \frac{c^2(4n^3-n)}{3} = \frac{n(4n^2-1)c^2}{3}}$

$\therefore \mathbf C$ is the right option.


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