find the sum of all integers between 100 and 200 that are divisible by 9.
Answers
Answered by
4
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
Answered by
0
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
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