find the sum of the first 40 posative integers divisible by 6
Answers
Answered by
1
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
Answered by
0
Hello guys you can check my answer , but I am not fully sure about the answer .
Attachments:
Similar questions