Average of first $10$ multiples of $3$ is:
Average of first $10$ multiples of $3$ is:
- $15$
- $16.5$
- $18$
- $20$
Solution
First $10$ multiples of $3$: $3,6,9,\ldots,30$. Sum $= 3(1+2+\cdots+10) = 3 \cdot 55 = 165$. Average $= \dfrac{165}{10} = 16.5$.
Asked in: IMO
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