progresions, If the sum of first n terms of an A.P. is cn<sup>2</sup>, then sum of squares of these n terms is

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Title: progresions

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Posted On: Feb 13, 2012 11:17 AM

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)c²/3

HOW?

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santosh p
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Pavan Kumar

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Answer Details:

Feb 13, 2012 11:41 AM

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

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

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

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

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

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

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