Question

Find the sum of all $2-$digit no is which when divided by $3$ yield $1$ as remainder.

Answer 1

Two digits numbers which when divides by $3$ yield $1$ as remainder are $10,13,16,................97$

This is an A.P. with $a=10,d=3$

Let $an=97$

$\Rightarrow a+\left(n-1\right)d=97$

$\Rightarrow 10+\left(n-1\right)3=97$

$\Rightarrow 10+3n-3=97$

$\Rightarrow 3n+7=97$

$\Rightarrow 3n=90$

$\Rightarrow n=30$

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

$=\frac{30}{2}\left[2\times 10+\left(30-1\right)3\right]$

$=15\left[20+29\times 3\right]$

$=15\left[20+87\right]$

$=15\times 107$

$=1605$