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Previous Year Question (PYQs)

Let sum of $n$ terms of an A.P. is $2n(n-1)$, then sum of their squares is




Solution

Sum $S_n = 2n(n-1)$ → $a = 2$, $d = 4$ Sum of squares = $\sum (a + (r-1)d)^2$ $= n[a^2 + (n-1)(a+d)^2 + …]$ Simplifying gives $\dfrac{8n(n+1)(2n+1)}{3}$

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