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Find the sum of the integers between 100 and 200 that are not divisible by 9
Given:
To do:
We have to find the sum of the integers between 100 and 200 that are not divisible by 9.
Solution:
Sum of the integers between 100 and 200, which are not divisible by 9 $=$ Sum of the integers between 100 and 200 $-$ Sum of the integers between 100 and 200, which are divisible by 9
Integers between 100 and 200 are \( 101,102,103, \ldots, 199 \).
The sequence is in A.P.
Here,
\( a=101 \) and \( d=102-101=1 \) \( l=199 \)
We know that,
$l=a+(n-1) d$
$\Rightarrow 199=101+(n-1) \times 1$
$\Rightarrow 199=101+n-1$
$\Rightarrow 199-100=n$
$\Rightarrow n=99$
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$=\frac{99}{2}[2 \times 101+(99-1) \times 1]$
$=\frac{99}{2}[202+98 \times 1]$
$=\frac{99}{2}(202+98)$
$=99 \times 150$
$=14850$
Integers between 100 and 200, which are divisible by 9 are \( 108,117,126, \ldots, 198 \).
The sequence is in A.P.
Here,
\( a=108 \) and \( d=117-108=9 \)
\( l=198 \)
We know that,
$l=a+(n-1) d$
$\Rightarrow 198=108+(n-1) \times 9$
$\Rightarrow 198=108+9n-9$
$\Rightarrow 198-99=9 n$
$\Rightarrow n=\frac{99}{9}=11$
Therefore,
$n=11$
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$=\frac{11}{2}[2 \times 108+(11-1) \times 9]$
$=11[108+5 \times 9]$
$=11(108+45)$
$=11 \times 153$
$=1683$
Therefore,
Sum of all
$=13167$
The sum of all integers between 100 and 200 which are not divisible by 9 is $13167$.