If $S_{n} = 2 n^{2} + 3 n$ then $d = . . . . . . .$

13

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4

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9

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- 2

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Solution

The correct option is B $4$
$S_{n} = 2 n^{2} + 3 n$
$S_{n - 1} = 2 {(n - 1)}^{2} + 3 (n - 1)$
$t_{n} = S_{n} - S_{n - 1}$
$= 2 n^{2} + 3 n - 2 {(n - 1)}^{2} - 3 (n - 1)$
$= 2 (n^{2} - {(n - 1)}^{2}) + 3 (n - n + 1)$
$= 2 (n - n + 1) (n + n - 1) + 3$
$= 2 (2 n - 1) + 3$
$= 4 n - 2 + 3 = 4 n + 1$
$t_{1} = 4 \times 1 + 1 = 4 + 1 = 5$
$t_{2} = 4 \times 2 + 1 = 8 + 1 = 9$
Common difference $= d = t_{2} - t_{1} = 9 - 5 = 4$

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