Find the sum of first 40 positive integers divisible by 6
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1
The first 40 positive integers that are divisible by 6 are 6,12,18,24…
a=6 and d=6.
We need to find S40
Sn=2n[2a+(n−1)d]
S40=240[2(6)+(40−1)6]
=20[12+(39)6]
=20(12+234)
=20×246
=4920
Answered by
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Step-by-step explanation:
positive integers divisible by 6 are
6,12,18,24,....
since difference is same, it is an AP
we need to find sum of first 40 integers
we can use formula
Sn=n/2(2a+(n-1)d)
here n=40
a=6 & d=12-6=6
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