The sum of 'r' terms of an AP is $S_{r} = (2 r^{2} + 3 r)$ . The $n^{t h}$ term is :

2n - 1

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4n + 1

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3n - 1

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3n + 1

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Solution

The correct option is B
4n + 1

$S_{r} = (2 r^{2} + 3 r)$
$T_{n} = S_{n} - S_{n - 1}$
So, $S_{n} = 2 n^{2} + 3 n$
$S_{n - 1} = 2 {(n - 1)}^{2} + 3 (n - 1)$
Now,
$T_{n} = S_{n} - S_{n - 1}$
$= 2 n^{2} + 3 n - [2 (n^{2} - 2 n + 1) + 3 n - 3]$
$= 2 n^{2} + 3 n - 2 n^{2} + 4 n - 2 - 3 n + 3 = 4 n + 1$

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