Question

Find the sum of all two digit numbers, which when divided by 4, yields 1 as remainder.

Answer 1

The two-digit numbers which when divided by 4 yield 1 as remainder are 13, 17, …… 97.
$\therefore a=13,d=4,a_n=97\therefore a_n=a+(n-1)d\Rightarrow 97=13+(n-1)4\Rightarrow 84=4n-4\Rightarrow 88=4n\Rightarrow 22=n\dots \dots \left(i\right)$
Also, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$S_{22}=\frac{22}{2}[2\times 13+(22-1)\times 4]$ [From (i)]
$\Rightarrow S_{22}=11\left[110\right]=1210$