Find the sum of first 40 positive integer divisible by 6
Answer (1)
22nd Nov, 2024
We can achieve this by using the formula for the sum of an arithmetic series:
Sn = n/2 * (2a + (n-1)d)
where
Sn = Sum of n terms
n = Number of terms =40
a = First term =6
d = Common difference =6
We substitute the values:
Sn = 40/2 * (2*6 + (40-1)6)
= 20 * (12+396)
= 20 * (12+234)
= 20 * 246
= 4920
Therefore, the sum of the first 40 positive integers that are divisible by 6 is 4920.
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