Question

find the sum of all integers between 100 and 200 that are divisible by 9.

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Answer 1

Answer:

Sum :- 1683

Step-by-step explanation:

⚠️ Answer ⚠️

Given,

a = 108

d = 9

an = 198

First Find :- Number of terms

=> an = a + (n-1)d

=> 198 = 108 + (n-1)9

=> 198 - 108 = 9n - 9

=> 90 + 9 => 9n

=> 9n = 99

=> n = 11

Now Find :- Sum of terms

=> Sn = n/2 [ 2a + (n-1)d]

=> Sn = 11/2 [ 2x108 + (11 - 1)9]

=> Sn = 11/2 ( 306 )

=> Sn = 11 x 153

=> Sn = 1683

Answer 2

Answer:

1683

Step-by-step explanation:

Simplest solution:>

number less and near200 divisible by 9

200 - remainder of 200/9

= 200 - 2= 198 = An

number greater and near 100 divisible by 9

100 - remainder of 100/9 +9

100-1+9= 108 = A1

number of terms= ( An - A1) /d+1

= (198 - 108) / 9+1

thus n = 90/9+1= 11

sum = (n/2) (A1+An)

= (11/2) (198+108)= 306×11/2

= 1683