Question

# Find the sum of all integers between 100 and 550, which are divisible by 9.

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Solution

## The integers between 100 and 550 that are divisible by 9 are: 108, 117...549 Here, we have: $a=108\phantom{\rule{0ex}{0ex}}d=9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{n}=549\phantom{\rule{0ex}{0ex}}⇒108+\left(n-1\right)\left(9\right)=549\phantom{\rule{0ex}{0ex}}⇒9n-9=441\phantom{\rule{0ex}{0ex}}⇒9n=450\phantom{\rule{0ex}{0ex}}⇒n=50\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}⇒{S}_{50}=\frac{50}{2}\left[2×108+\left(50-1\right)×9\right]\phantom{\rule{0ex}{0ex}}⇒{S}_{50}=25\left(657\right)=16425\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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