Question

find the sum of integers from 100 to 200 that are divisible by 9 ​

Answer 1

108,117,126.... 198 and so on

So a=108

l=198

n= ?

nth term= a + (n-1)d

198= 108+(n-1)9

198= 108+9n-9

198-99= 9n

99/9= n

n= 11

Now, sum of n terms

Sum of first 11 terms= 11/2(108+198)

=11/2(306)

=11×153

=1683

Answer 2

108,117,126......................198.

this is an AP with

d=9

a=108

l=198=an

an=a+(n-1)d

198=108+(n-1)9

10=n-1

n=11

sum of integer=n(a+l)/2=11(108+198//2

=11(306)/2=153×11=1683

hope this will help u......