If the sum of $n$ terms of an AP is $2 n^{2} + 5 n,$ then its $n t h$ term will be

4 n - 3

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3 n - 4

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4 n + 3

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3 n + 4

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Solution

The correct option is A $4 n + 3$
Given, $S_{n} = 2 n^{2} + 5 n$

$\Rightarrow S_{1} = 2 (1)^{2} + 5 (1) = 7 = a$

$S_{2} = 2 (2)^{2} + 5 (2) = 18 = a_{1} + a_{2}$

$a_{1} + a_{2} = 18$

$\Rightarrow 7 + a_{2} = 18$

$\Rightarrow a_{2} = 11$

$∴ d = a_{2} - a_{1} = 11 - 7 = 4$

Since, $a_{n} = a + (n - 1) d$

$\Rightarrow a_{n} = 7 + (n - 1) 4 = 7 + 4 n - 4 = 4 n + 3$
or
$S_{n} - S_{n - 1} = a_{n}$
$S_{n} = 2 n^{2} + 5 n$
$S_{n - 1} = 2 (n - 1)^{2} + 5 (n - 1)$
$S_{n} - s_{n - 1} = 2 n^{2} + 5 n - 2 (n - 1)^{2} - 5 (n - 1)$
$S_{n} - S_{n - 1} = a_{n} = 4 n + 3$

$∴$ Option C is correct.

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