Find the sum of all the integers between 100 and 200 that are not divisible by 9

Here this is the sequence of an A.P. where first term;

*a*= 108, common difference;

*d*= 9

Here nth term of an A.P. is 198, so we have;

${t}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\Rightarrow 198=108+\left(n-1\right)\times 9\phantom{\rule{0ex}{0ex}}\Rightarrow 9\left(n-1\right)=198-108\phantom{\rule{0ex}{0ex}}\Rightarrow 9\left(n-1\right)=90\phantom{\rule{0ex}{0ex}}\Rightarrow \left(n-1\right)=10\phantom{\rule{0ex}{0ex}}\Rightarrow n=10+1=11$

So sum of 11 terms is given by;

108 + 117 + 126 + … + 198 = $\frac{n}{2}\left[2a+\left(n-1\right)d\right]=\frac{11}{2}\left[2\times 108+\left(11-1\right)\times 9\right]=1683$

So the sum of integers between 100 and 200 which are divisible by 9 is 1683.

The sum of the integers between 100 and 200 which are not divisible by 9

= The sum of all integers between 100 and 200 – The sum of the integers between 100 and 200 which are divisible by 9

$=\left(101+102+103+...+199\right)-1683\phantom{\rule{0ex}{0ex}}=\left[\left(1+2+3+4+...+199\right)-\left(1+2+3+4+..100\right)\right]-1683\phantom{\rule{0ex}{0ex}}=\frac{199\times 200}{2}-\frac{100\times 101}{2}-1683\phantom{\rule{0ex}{0ex}}=19900-5050-1683\{1+2+3+...+n=\frac{n\left(n+1\right)}{2}\}\phantom{\rule{0ex}{0ex}}=13167$

Therefore the sum of the integers between 100 and 200 which are not divisible by 9 is 13167.

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