find the sum of the natural numbers between 101 and 999, which are divisible by both 2 and 5
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The first number after 101 divisible by 2 and 5 is 110.
So,
a = 110,
d = 10 (because 2nd term will be 120),
l (which is last term) = 990.
First we have to find the no : of terms between 101 and 999 which are divisible both by 2 and 5. For that the equation is :
l = a + (n-1) x d
990 = 110 +(n-1) x 10
990 - 110 = (n-1) x 10
880/10 = n-1
88 = n-1
n = 89
∴ There are 89 terms between 101 and 999 which are divisible both by 2 and 5.
Now finding their sum,
Sn = n/2 [2a (n-1) x d]
Sn = 89/2 [2 x 110 x 88 x 10]
Sn = 44.5 [193600]
Sn = 8615200
∴ The sum of the natural numbers between 101 and 999, which are divisible both by 2 and 5 is 8615200
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