Question

Find the sum of the AP: $50,46,42,\dots$upto $10$ terms.

Answer 1

Find the sum of the AP up to $10$ terms:

We know that the sum of first $n$ terms of an AP is

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

Where $a=$ first term and $d=$ common difference

We have, AP: $50,46,42,\dots$upto $10$ terms

Here, first term $a=50$

common difference $d=46-50=-4$ and $n=10$

$\begin{array}{rcl}\therefore S_{10}& =& \frac{10}{2}[2\times 50+(10-1)\times (-4\left)\right]\\ & =& 5[100-9\times 4]\\ & =& 5[100-36]\\ & =& 5\times 64\\ & =& 320\end{array}$

Hence, the required answer is $320$.