Find the sum of all integers between 100 and 600, each of
Question:

Find the sum of all integers between 100 and 600, each of which when divided by 5 leaves 2 as remainder.

Solution:

The integers between 100 and 600 divisible by 5 and leaves remainder 2 are 102, 107, 112, 117,…, 597.

To Find: Sum of the above AP

Here a = 102, d = 5, l = 597

$a+(n-1) d=597$

$\Rightarrow 102+5(n-1)=597$

$\Rightarrow(n-1)=99$

$\Rightarrow n=100$

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

$\Rightarrow S=\frac{100}{2}[2 \times 102+5(100-1)]$

$\Rightarrow S=50[204+495]=50 \times 699=34950$

The sum of all such integers is 34950.