Question

Question 5 (i)
Find the sum of the integers between 100 and 200 that are divisible by 9.

Answer 1

The numbers (integers) between 100 and 200 which is divisible by 9 are 108, 117, 126, ……….. 198.
Let n be the number of terms between 100 and 200 which is divisible by 9.
Here, a = 108, d = 117 – 108 = 9 and $a_n$ = l = 198
$\because a_n=l=a+(n-1)d$
198 = 108 + (n - 1)9
10 = n - 1
n = 11
$\therefore$ Sum of terms between 100 and 200 which is divisible by 9,
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow S_{11}=\frac{11}{2}\left[2\right(108)+(11-1\left)9\right]=\frac{11}{2}[216+90]$
$=\frac{11}{2}\times 306=11\times 153=1683$
Hence, required sum of the integers between 100 and 200 that are divisible by 9 is 1693.