What is the average of the first 92 even numbers?

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Accepted Answer

**Solution:**

1. The first 92 even numbers form an arithmetic progression (AP) with first term $a_1 = 2$ and common difference $d = 2$.

2. The $n^{th}$ even number is given by:
   $$a_n = 2n$$
   So, the 92nd even number is:
   $$a_{92} = 2 	imes 92 = 184$$

3. The average of the first $n$ even numbers is the average of the first and the last term of the AP:
   $$	ext{Average} = \frac{a_1 + a_n}{2} = \frac{2 + 184}{2} = \frac{186}{2} = 93$$

4. Alternatively, the formula for the average of first $n$ even numbers is:
   $$	ext{Average} = n + 1$$
   Substituting $n=92$:
   $$	ext{Average} = 92 + 1 = 93$$

**Answer:** $93$

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