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If the sum to n terms of an A.P is cn(n –1); c  0, then the sum of squares of these terms is

(a) c2 n2 (n + 1)2 

(b)  2c2/3 n(n - 1)(n - 1)

(c)  2c2/3n(n + 1)(2n + 1)

(d)  None of thes

1 Answer

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Best answer

Correct option (b)  2c2/3  n(n - 1)(n - 1)

Explanation:

Tn = Sn –Sn–1 = cn(n–1) –c(n–1)(n–2)

= c(n–1){n–n+2}

= 2c(n–1)

Tn = 4c2 (n–1)2

∴ Sn2 = 4c2{0+ 1+ 2+..................+ (n – 1)2}

=  4c2 n(n- 1)(2n - 1)/6

=  2c2/3  n(n - 1)(n - 1)

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