Question

Find the sum of last ten terms of the $A.P.:8,10,12,......,126.$

Answer 1

The sum of last ten terms of the AP: $8,10,12,...,126$ is same as the sum of first ten terms of the AP: $126,124,122,...,8$.
Therefore, $a=126,d=-2$ and $n=10$.
Now, $S_{10}=\frac{10}{2}[2\times 126+(10-1)\times -2]$
$\Rightarrow S_{10}=5\times [252-18]=1170$