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# If the sum of first n terms of an A.P. is cn2, then sum of squares of these n terms is?ans:n(4n2-1)c2/3HOW?

Pavan Kumar
39 Points
9 years ago

If a is 1st term and d is common difference, sum of n terms = (n/2) (2a + (n-1)d) = (2an-nd) + n2d

comparing this with cn2 we get : 2an-nd = 0  and  d = c    =>    a = c/2,  d = c

sum of squares of n terms = $\sum_{r=1}^{n}(a+(n-1)d)^{2}$

$=\sum_{r=1}^{n}\left( \frac{c}{2}+(n-1)c\right)^{2}$

$=\sum_{r=1}^{n}c^{2}\left( n^{2}-n+\frac{1}{4}\right)$

$=c^{2}\left( \sum_{r=1}^{n}n^{2}-\sum_{r=1}^{n}n+\sum_{r=1}^{n}\frac{1}{4}\right)$

$=c^{2}\left(\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+\frac{n}{4}\right)$

$=\frac{c^{2}n(4n^{2}-1)}{12}$

one year ago
Dear student,

If a is 1st term and d is common difference, sum of n terms = (n/2) (2a + (n – 1)d) = (2an – nd) + n2d
comparing this with cn2 we get : 2an-nd = 0  and  d = c    =>    a = c/2,  d = c
sum of squares of n terms = Σ(a + (n – 1)d)2
= Σ(c/2 + (n – 1)c)2
= c2 Σ(n2 – n + 1/4)
= c2n(4n2 – 1)/12

Thanks and regards,
Kushagra