if the sum of first n terms of an A.P. is cn2 ,then the sum of squares of these n terms is (A) n(4n2 -1)c2/2(B) n(4n2 +1)c2/3(C) n(4n2 -1)c2/3(D) n(4n2 +1)c2/6

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Grade 12Algebra

if the sum of first n terms of an A.P. is cn² ,then the sum of squares of these n terms is
(A) n(4n² -1)c²/2
(B) n(4n² +1)c²/3
(C) n(4n² -1)c²/3
(D) n(4n² +1)c²/6

Ankit Jaiswal

10 Years agoGrade 12

2 Answers

Riddhish Bhalodia

Approved Tutor Answer10 Years ago

A be the first term and d be the difference of the AP
sum of n terms is
n(A + d(n-1)/2) = cn²
simplifying we get the equation as
(A-d/2) + nd/2 = cn
now comparing coefficients in ‘n’ we get two relations as
A - d/2 = 0 and d/2 = c
thus A = c and d = 2c
then the AP (in terms of c and n) is as follows
c , 3c , 5c , 7c , ….. , (2n-1)c
Now we need to evaluate the sum of squares of these n terms i.e as
S = (1²+ 3²+ …. + (2n-1)²) c2
using the formula for the sum of squares of first n odd numbers we get the final value of S as
$S = \frac{c^2(2n-1)(2n+1)n}{3} = \frac{c^2n (4n^2-1)}{3}$
Hence the option C is correct answer