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Can anyone please tell me what would be the fastest way to find the average of all numbers from $1$ to $100$ that end in $2$?

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Evidently, the fastest way is to ask on this website. 2 answers so far in under 4 minutes. –  Gerry Myerson May 22 '12 at 1:40
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2 Answers

The numbers are of the form $10k+2$ where $k \in \{0,1,2,\ldots,9\}$. Hence, the average is $$\frac{\displaystyle \sum_{k=0}^{9} (10k+2)}{\displaystyle \sum_{k=0}^{9} 1} = 47$$ A quick way is to notice that the numbers are in arithmetic progression and hence the average is $$\dfrac{\text{First term}+\text{Last term}}{2} = \dfrac{2+92}{2} = 47$$

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Thanks that helped. Forgot about AP formula –  Rajeshwar May 22 '12 at 1:42
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Group $2$ with $92$, $12$ with $82$, $22$ with $72$, $32$ with $62$, and $42$ with $52$. Each group can be further "split" into $47$ and $47$ (for example $2+92=47+47$). So the average is $47$.

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