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In the sum of first n terms of an A.P. i...

In the sum of first n terms of an A.P. is `cn^2`, then the sum of squares of these n terms is

A

`(n(4n^(2) -1)c^(2))/(6)`

B

`(n(4n^(2) + 1)c^(2))/(3)`

C

`(n(4n^(2) - 1)c^(2))/(3)`

D

`(n(4n^(2) + 1)c^(2))/(6)`

Text Solution

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The correct Answer is:
C
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Knowledge Check

  • If the sum of first n terms of an AP is cn^(2) , then the sum of squares of these n terms is

    A
    `(n(4n^(2)-1)c^(2))/(6)`
    B
    `(n(4n^(2)+1)c^(2))/(3)`
    C
    `(n(4n^(2)-1)c^(2))/(3)`
    D
    `(n(4n^(2)+1)c^(2))/(6)`

  • If the sum of first n terms of an AP is "cn"^(2) , then the sum of squares of these n terms is

    A
    `(n (4n^(2) - 1)c^(2))/(6)`
    B
    `(n (4n^(2) + 1)c^(2))/(3)`
    C
    `(n (4n^(2) - 1)c^(2))/(3)`
    D
    `(n(4n^(2) + 1)c^(2))/(6)`

  • If the sum of n terms of an A.P. is an(n - 1), then sum of squares of these n terms is

    A
    `(2a^(2))/(3) n (n - 1) (2n - 1)`
    B
    `(2a^(2))/(3) n(n + 1) (2n + 1)`
    C
    `(a^(2))/(6) n (n - 1) (2n - 1)`
    D
    `(a^(2))/(6) n (n + 1) (2n + 1)`

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