The average of 5 consecutive integers is 10. What is the smallest integer?
8
Let the 5 consecutive integers be \(n, n+1, n+2, n+3, n+4\).
Their average is \(\frac{n + (n+1) + (n+2) + (n+3) + (n+4)}{5} = \frac{5n + 10}{5} = n + 2\).
Setting \(n + 2 = 10\) gives \(n = 8\).
Hence, the smallest of the 5 consecutive integers is 8.
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Important Questions from Average
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