Find the sum of all those integers between 100 and 800

Solution:

The sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7 are:

103, 119...791

Here, we have:

a = 103

d = 16

$a_{n}=791$

We know:

$a_{n}=a+(n-1) d$

$\Rightarrow 791=103+(n-1) \times 16$

$\Rightarrow 688=16 n-16$

$\Rightarrow 704=16 n$

$\Rightarrow 44=n$

Also, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow S_{44}=\frac{44}{2}[2 \times 103+(44-1) \times 16]$

$\Rightarrow S_{44}=22[206+688]$

$\Rightarrow S_{44}=22 \times 894=19668$