Question

Find the sum of all two-digit numbers which, being divided by 4, leave a remainder of 1.

Answer 1

The two-digit numbers, which when divided by $4$, yield $1$ as remainder, are

$13,17,\dots 97$.

This series forms an A.P. with first term $13$ and common difference $4$.

Let $n$ be the number of terms of the A.P.

It is known that the $n^{th}$ term of an A.P. is given by,

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

$97=13+(n-1)4$

$4(n-1)=84$

$n-1=21$

$\Rightarrow n=22$

Sum of the numbers is:

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

$S_n=\frac{22}{2}(2\times 13+(22-1\left)4\right)$

$S_n=1210$