If the sum of n terms of an A-P- is c nn-1- where k-0- then sum of the squares of these terms is

The correct option is C If t_r nbsp;be the r^th nbsp;term of the A.P., then t_r=S_r S_r 1 cr(r 1) c(c 1)(r 2) =c(r 1)(r r+2)=2c(r 1) we have, t_1^2+r_2^2+ ...

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Standard XII

Mathematics

Sigma n2

If the sum of...

Question

If the sum of n terms of an A.P. is cn(n-1), where c $\neq$ 0, then sum of the squares of these terms is

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none of these

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Solution

The correct option is B

If $t_{r}$ be the $r^{t h}$ term of the A.P., then

$t_{r} = S_{r} - S_{r - 1} c r (r - 1) - c (c - 1) (r - 2)$

$= c (r - 1) (r - r + 2) = 2 c (r - 1)$
we have,
$t_{1}^{2} + r_{2}^{2} + . . . . . . + t_{n}^{2} = 4 c^{2} (0^{2} + 1^{2} + 2^{2} + . . . . . + (n - 1)^{2})$
$= 4 c^{2} \frac{(n - 1) n (2 n - 1)}{6}$
$= \frac{2}{3} c^{2} n (n - 1) (2 n - 1)$

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If the sum of n terms of an A.P. is cn(n-1), where c≠0, then sum of the squares of these terms is

The correct option is B If tr be the rth term of the A.P., then tr=Sr−Sr−1cr(r−1)−c(c−1)(r−2) =c(r−1)(r−r+2)=2c(r−1) we have, t21+r22+......+t2n=4c2(02+12+22+.....+(n−1)2) =4c2(n−1)n(2n−1)6 =23c2n(n−1)(2n−1)

If the sum of n terms of an A.P. is cn(n-1), where c $\neq$ 0, then sum of the squares of these terms is