Find the sum of the following pattern:$30+31+32+.......+50$

Given :

Given series is $30+31+32+.....+50$

To find :

We have to find the sum of the given terms.

Solution :

First term = a = 30

Common difference $= 31-30 = 1$.

$a_{n}=50$

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

$50 = 30+(n-1)1$

$50-30=n-1$

$n = 20+1 = 21$

Sum of 'n' terms :

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

     $ = \frac{21}{2} [2\times30+(21-1)1]$

     $ = \frac{21}{2} [60+20]$

     $ = \frac{21}{2} \times 80$

    $= 21\times 40$

      $= 840$.

Therefore, Sum of $30+31+32+.....+50$ is  $840$.

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Updated on: 10-Oct-2022

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