Question

find the sum of the integers between 100 and 200 that are divisibke by 9

Answer 1

the series is given by,

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

n=11

Now, sum of n terms

Sum of first 11 terms

=

2

11

(108+198)

=

2

11

(306)

=11×153

=1683

HOPE IT HELPS ☺️