Question

The sum of integers between 100 to 200 divisible by 9

Answer 1

Answer:

Step-by-step explanation:

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

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