Question

2. Find the sum of all 3 digit numbers
that are divisible by 4.
​

Answer 1

3 digit numbers that are divisible by 4 are 100,104,108...996

first term (a) = 100

common difference (d) = 4

last term (l) = 996

$Total \: number \: of \: terms = \frac{996 - 100}{4} + 1 \\ = \frac{896}{4} + 1 \\ = 224 + 1 \\ = 225$

Therefore, there are total 225 numbers of 3 digit which are divisible by 4.

$using \: fomula \: of \: s_{n}(sum \: of \: n \: terms) \\ sum \: of \: all \: three \: digit \: numbers \: which \: are \: divisible \: by \: 4 \: = \frac{225}{2} (100 + 996) = 123,300$