Find the sum of all those integers between 100 and 800

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

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

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

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