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If the sum of n terms of an A.P, is cn (n-1) , where , then sum of the squares of these terms is
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a
c2nn+12
b
23c2nn−12n−1
c
2c23nn+12n+1
d
c2
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Correct option is B
if tr be the rth term of the A.P. then tr=Sr−Sr−1=cr(r−1)−c(r−1)(r−2)=c(r−1)(r−r+2)=2c(r−1) We have, t12+t22+….+tn2=4c202+12+22+….+(n−1)2=4c2(n−1)n(2n−1)6= 23c2n(n−1)(2n−1)Similar Questions
Let be a positive integer, then is equal to :
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