Find the sum of the first 40 positive integers divisible by 6
Answers
Answered by
2
Answer:
The first 40 positive integers that are divisible by 6 are 6,12,18,24…
a=6 and d=6.
We need to find S
40
S
n
=
2
n
[2a+(n−1)d]
S
40
=
2
40
[2(6)+(40−1)6]
=20[12+(39)6]
=20(12+234)
=20×246
=4920
Step-by-step explanation:
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Answered by
0
Answer:
6
Step-by-step explanation:
n=40. , a=6.
d=12-6=6
the answer came
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