Question

Find the sum of the ontegers between 100 and 200 thar are divisible by 9

Answer 1

First no a=108
last no l=198
d=9
total no tn=a+(n-1)d
198=108+9n-9
9n=198-99
n=99/9
n=11
sum of 11 no is
sn=n/2(a+l)
=11/2(108+198)
=11/2×306
=11×153
sn=1683

Answer 2

First term = a = 108
Last term = l = 198





$\boxed{ \bold{a_{n} = a + (n - 1)d}}$

=> 198 = 108 + ( n - 1 ) 9
=> 198 - 108 = ( n - 1 ) 9
=> 90 = ( n - 1 ) 9
=> 10 = n - 1
=> 11 = n


Hence, Number of term = n = 11




$\boxed{ \bold{ we \: know \: \: \: \: sum \: on \: n \: terms = \frac{n}{2} (a + l)}}$



So,

$s_{11} = \frac{11}{2} (a + l) \\ \\ \\ \\ s_{11} = \frac{11}{2} (108 + 198) \\ \\ \\ \\ s_{11} = \frac{11}{2} \times 306 \\ \\ \\ \\ s_{11} = 11 \times 153 \\ \\ \\ \\ s_{11} = 1683$