Find the sum of given series
$31 + 32 + 33 + . . . . . + 50$

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Solution

Given,

$a = 31$

$n = 20$

$d = 32 - 31 = 1$

Formula:

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$= \frac{20}{2} [2 (31) + (20 - 1) 1]$

$= 10 [81]$

$= 810$

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