If
the sum of first n terms of A.P is cn

^{2}, then the sum of squares of these n terms is:
(a)n(4n

^{2}-1)c^{2}/6 (b) n(4n^{2}+1)c^{2}/3 (c) n(4n^{2}-1)c^{2}/3 (d) n(4n^{2}+1)c^{2}/6
Solution
follows here:

__Solution:__
Sum of first n
terms = S

_{n}= cn^{2}
∴Sum of first (n-1) terms = S

_{n-1}= c(n-1)^{2}
n th term = t

_{n}= S_{n}- S_{n-1}= c{n^{2}– (n-1)^{2}} = c(2n-1)
To find sum of
squares of first n terms of this series, we need n th term of the series of
squares of the respective terms.

n th term of
the series of squares of the respective terms = t

_{n}^{’}= {c(2n-1)}^{2}= c^{2}(2n-1)^{2}
Sum of n terms
of the series of squares of the respective terms

= S

_{n}^{’}= ∑c^{2}(2n-1)^{2}= c^{2}∑(2n-1)^{2}= c^{2}∑(4n^{2}-4n+1) = c^{2}{4∑n^{2}-4∑n+∑1}
= c

^{2}{4n(n+1)(2n+1)/6 – 4n(n+1)/2 + n}
= n(4n

^{2}-1)c^{2}/3**Answer (c)**

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