If the sum of first n terms of an A.P. is cn^{2}, then the sum of squares of these n terms is: Sequence & Series (Mathematics) Question | DivineJEE

If the sum of first

n

terms of an A.P. is

cn^{2}

, then the sum\n of squares of these

n

terms is

\frac{n\left( 4n^{2} - 1 \right)c^{2}}{6}

\frac{n\left( 4n^{2} + 1 \right)c^{2}}{3}

\frac{n\left( 4n^{2} - 1 \right)c^{2}}{3}

\frac{n\left( 4n^{2} + 1 \right)c^{2}}{6}

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