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.
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