The sum of the first 10 natural numbers which are divisible by 3
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Answered by
3
Answer:
3+6+9+12+15+18+21+24+27+30 = 165
Answered by
2
Answer:
sequence is 3,6,9,......,30
a = 3
d = 6-3 = 3
n = 10
Sn = n/2(2a+(n-1)d
S10 = 10/2(2(3)+(9)3)
= 5(6+27)
= 5(33)
= 165
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