If the sum of first n terms of an A.P. is cn( | vedclass

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Question

$\mathrm{S}_{\mathrm{n}}=\mathrm{cn}(\mathrm{n}-1)$

$S_{n-1}=c(n-1)(n-2)$

$t_{n}=S_{n}-S_{n}=c(n-1)(n-(n-2))=2 c(n-1)$

So, for the new series

Vedclass · Accepted Answer

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

Vedclass · Answer

$\mathrm{S}_{\mathrm{n}}=\mathrm{cn}(\mathrm{n}-1)$

$S_{n-1}=c(n-1)(n-2)$

$t_{n}=S_{n}-S_{n}=c(n-1)(n-(n-2))=2 c(n-1)$

So, for the new series

$\mathrm{T}_{\mathrm{n}}=4 \mathrm{c}^{2}(\mathrm{n}-1)^{2}=4 \mathrm{c}^{2}\left[\mathrm{n}^{2}-2 \mathrm{n}+1\right]$

So,${S_n}$  for the new series will give

${S_n} = 4{c^2}\left[ {\sum {{n^2} - 2\sum {n + n} } } \right] = \frac{2}{3}{c^2}n(n - 1)(2n - 1)$

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

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