Question

Find the sum of all two digit numbers greater than $50$ which when divided by $7$, leave a remainder of $4$.

Answer 1

The number should be the form $7n+4$

Should be $50<7n+4<100$

$\therefore$ Numbers are $53,60,\dots ..95$

$a=53$

$d=7$

$n^{th}$ term $=a+(n-1)d$

$\Rightarrow 53+(n-1)7=95$

$\Rightarrow (n-1)\text{/}7={\text{/}42}^6$

$\Rightarrow n=7$

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

$=\frac{7}{2}(2\times 53+(7-1\left)7\right)$

$=\frac{7}{2}[106+42]$

$=518$.