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What is the average of all the one digit, two digit and three digit natural numbers?
- A300
- B500
- C1000
- D1250
Answer not set
Discussion 1
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CCommunity
As we know, all the one digit, two digit, and 3-digit natural numbers are from 1 – 999
? Sum of n natural numbers = n(n + 1)/2
Sum of 999 natural numbers = (999 × 1000)/2 = 499500
∴ Required average = 499500/999 = 500
? Sum of n natural numbers = n(n + 1)/2
Sum of 999 natural numbers = (999 × 1000)/2 = 499500
∴ Required average = 499500/999 = 500