Question

Find the sum of the first $50$ even positive integers.

Answer 1

Sum of first $n$ terms, $S_n=\frac{n}{2}(2a+(n-1\left)d\right)$, where $a$ & $d$ are the first term & common difference of an AP.

The sequence of the first $50$ even positive integers is given by $2,4,6,...$

The above sequence has a first term $a=2$ and a common difference $d=2$.

And $n=50$

So,the sum of the first $50$ terms

$S_{50}=\frac{50}{2}(2\times 2+(50-1)\times 2)$

$\Rightarrow S_{50}=25(4+49\times 2)$

$\Rightarrow S_{50}=25(4+98)$

$\Rightarrow S_{50}=2550$