Question

Find the sum of the series: $31,32,....,50$

Answer 1

Given sequence is $31,32,....,50$

It is an AP with first term $a=31$

and, common difference $d=32-31=1$

Let, the number of terms be $n$

We know that the $n^{th}$ term,

$a_n$ $=a+(n-1)d$

$\Rightarrow 50=31+(n-1)\times 1$

$\Rightarrow 50-31=n-1$

$\Rightarrow 50-31+1=n$

$\Rightarrow n=20$

Therefore, the sum of terms,

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

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

$=10[62+19]$

$=810$

Hence, sum of given sequence is $810$.