Find the sum of n terms of the series $(a^{2} + b) + (a^{2} + 2 b) + (a^{2} + 3 b) + . . . .$

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Solution

We have,
A.P. is
$(a^{2} + b) + (a^{2} + 2 b) + (a^{2} + 3 b) . . . . . . . .$
First term $A = (a^{2} + b)$
Common difference
$d = b$
Then,
Sum of n terms
$S_{n} = \frac{n}{2} (2 A + (n - 1) d)$
$= \frac{n}{2} [2 \times (a^{2} + b) + (n - 1) b]$
$= \frac{n}{2} [2 a^{2} + 2 b + n b - b]$
Hence, this is the answer.

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