Find the sum of the first 40 integers divisible by 6
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1
Answer:
6,12,18,24,30,36................
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2
Answer:
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
Step-by-step explanation:
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