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
[2a+(n−1)d]
[2(6)+(40−1)6]
=20[12+(39)6]
=20(12+234)
=20×246
=4920
The first 40 positive integers that are divisible by 6 are 6,12,18,24...
here,
a=6
d=12-6=6
then,
Sn=n/2{2a+(n-1) d}
S40= 40/2{2×6+39×6}
= 20{12+234}
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Answers & Comments
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
The first 40 positive integers that are divisible by 6 are 6,12,18,24...
here,
a=6
d=12-6=6
then,
Sn=n/2{2a+(n-1) d}
S40= 40/2{2×6+39×6}
= 20{12+234}
=20×246
=4920