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In the sum of first n terms of an A.P. is `cn^2`, then the sum of squares of these n terms isA. `(n(4n^(2)-1))/(6)c^(2)`B. `(n(4n^(2)+1))/(3)c^(2)`C. `(n(4n^(2)-1))/(3)c^(2)`D. `(n(4n^(2)+1))/(6)c^(2)` |
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Answer» Correct Answer - C We have, `S_(n)=cn^(2)` `:.` First term = c and common difference d=2c So, given A.P. is c,3c,5c,7c, . . . . . . Let S be the sum of first n terms of A.P. then, `S=underset(r=1)overset(n)sum(2r-1)^(2)c^(2)` `rArr" "S=c^(2)underset(r=1)overset(n)sum(4r^(2)-4r+1)` `rArr" "S=c^(2){(4n(n+1)(2n+1))/(6)-(4n(n+1))/(2)+n}=(n(4n^(2)-1))/(3)c^(2)` ALITER We have, `S_(n)=cn^(2)` `:." "a_(n)=S_(n)-S_(n-1)=cn^(2)-c(n-1)^(2)=(2n-1)c` Let S be the sum of n terms of the sequence. Then, `S=underset(r=1)overset(n)suma_(r)^(2)` `rArr" "S=underset(r=1)overset(n)sum(2r-1)^(2)c^(2)` `rArr" "S=c^(2)underset(r=1)overset(n)sum(4r^(2)-4r+1)` `rArr" "S=c^(2){(4n(n+1)(2n+1))/(6)-(4n(n+1))/(2)+n}=(n(4n^(2)-1))/(3)c^(2)` |
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