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9Â years, 7Â months ago
the average of the first nine integral multiples of 3 is
9Â years, 7Â months ago
The sum of the series is 3*(1+2+3+...+9) = 3 * 9 * 10/2
Required average = (3*9*10/2) / 9 = 3 * 10/2 = 15
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