Question

Question 5 (ii)
Find the sum of the integers between 100 and 200 that are not divisible by 9.

Answer 1

The sum of the integers between 100 and 200 which is not divisible by 9 = ( sum of total numbers between 100 and 200) – (sum of total numbers between 100 and 200 which is divisible by 9).
Here, a = 101, d =102 – 101 = 1 and $a_n$ = l = 199
$\because a_n=l=a+(n-1)d$
$\Rightarrow 199=101+(n-1)1$
$\Rightarrow (n-1)=98\Rightarrow n=99$
$\therefore$ Sum of terms between 100 and 200,
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow S_{99}=\frac{99}{2}\left[2\right(101)+(99-1\left)1\right]=\frac{99}{2}[202+98]$
$=\frac{99}{2}\times 300=99\times 150=14850$
Therefore, sum of the integers between 100 and 200 which is not divisible by 9,
= 14850 – 1683
= 13167
Hence, the required sum is 13167.