Question

find the sum of integers between 100 & 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

Step-by-step explanation:

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

now,

a = 108

d = 117 - 108 = 9

according to question,

therefore, An = 198

a + (n - 1 ) d = 198

108 + (n-1) 9 = 198

( n-1) × 9 = 198 - 108

n - 1 = 90 ÷ 9

n = 10 + 1

n = 11

hello i am sure it os the correct answer