What is the sum of three digit number that are divisible by 3.
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Answer:
Answer : n=3.sum of digits=3+4+5=12
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a+(n-1)d = nth term of an arithmetic progression
a = 102
d = 3
nth term = 333
102 + (n-1) × 3 = 333
102 + 3n - 3 = 333
34 + n - 1 = 111
n = 78
Sum of n terms of an arithmetic progression = (n/2) × (2a + nd - d)
= (78/2) × (2×102+78×3-3)
= 39×(204+234-3)
= 39×(438-3)
= 39×435
= 16965
:-)
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