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If the sum of first n terms of an A.P. is $$cn^2$$, then the sum of squares of these n terms is
$$\frac{n(4n^2 - 1)c^2}{6}$$
$$\frac{n(4n^2 + 1)c^2}{3}$$
$$\frac{n(4n^2 - 1)c^2}{3}$$
$$\frac{n(4n^2 + 1)c^2}{6}$$
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