Question 187310
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If the average of 9 numbers is 40, then the sum of the nine numbers must be 9 times 40 or 360.  If you add 1 to the first number, 2 to the second, and so on, then you must be adding:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sum_{i=1}^9\, i = \frac{(1 + 9)9}{2} = 45]


So the sum is 360 + 45 = 405, and the new average is 405 divided by 9 = 45.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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