Find the sum of first 40 positive integers divisible by 6.
Answers
Answered by
22
hi friend,
the first 40 positive integers divisible by 6 are 6,12,18,....... upto
40 terms
the given series is in arthimetic progression with first term a=6 and common difference d=6
sum of n terms of an A.p is
n/2×{2a+(n-1)d}
→required sum = 40/2×{2(6)+(39)6}
=20{12+234}
=20×246
=4920
I hope this will help u ;)
Answered by
2
Answer:
The positive integers that are divisible by 6 are
6, 12, 18, 24 …
It can be observed that these are making an A.P. whose first term is 6
and common difference is 6.
a = 6 and d = 6
sn=n/2(2a+(n-I) d)
S40 =?
= 20[12 + (39) (6)]
= 20(12 + 234)
= 20 × 246
= 4920
please mark my answer as brainlist.
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