Find the sum of all natural numbers upto 100 which are divisible by 3.
Answers
Answered by
27
Answer:
1683
Step-by-step explanation:
The integers from 1 to 100, which are divisible by 3, are 3,6,9,12...99.
This forms an A.P with both the first term and common difference as 3.
Let n = number of terms.
a(n) = a + (n - 1) * d.
99 = 3 + (n - 1) * 3
99 = 3 + 3n - 3
99 = 3n
n = 33.
Sum of n terms of an AP sn = (n/2)[2a + (n - 1) * d]
= (33/2)[2(3) + (33 - 1) * 3]
= 33/2[6 + 96]
= 33/2[102]
= 33 * 51
= 1683.
Therefore,sum of natural numbers less than 100 divisible by 3 is 1683.
Hope it helps!
Answered by
2
Answer:
Step-by-step explanation:
Similar questions