Given that $S_1$ is a series of five consecutive multiples of 4 whose sum is 100, and $S_2$ is a series of four consecutive even integers such that the second smallest number in $S_2$ is 6 less than the largest number in $S_1$, find the average of the series $S_2$.

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Accepted Answer

**Solution:**

1. Let the five consecutive multiples of 4 in series $S_1$ be:
   $$4n, 4(n+1), 4(n+2), 4(n+3), 4(n+4)$$

2. Their sum is given as 100:
   $$
   4n + 4(n+1) + 4(n+2) + 4(n+3) + 4(n+4) = 100
   $$

3. Simplify the sum:
   $$
   4(n + n+1 + n+2 + n+3 + n+4) = 100
   $$
   $$
   4(5n + 10) = 100
   $$
   $$
   20n + 40 = 100
   $$

4. Solve for $n$:
   $$
   20n = 60 \
   n = 3
   $$

5. So, the multiples are:
   $$4 	imes 3 = 12, 16, 20, 24, 28$$

6. The largest number in $S_1$ is 28.

7. The second smallest number in $S_2$ is 6 less than 28:
   $$
  $ 	\text{Second smallest in } S_2 = 28 - 6 = 22$
   $$

8. Let the four consecutive even integers in $S_2$ be:
   $$x, x+2, x+4, x+6$$

9. Since the second smallest number is $x+2$, we have:
   $$
   x + 2 = 22 \
   x = 20
   $$

10. The four numbers in $S_2$ are:
    $$20, 22, 24, 26$$

11. Calculate the average of $S_2$:
    $$
    $\frac{20 + 22 + 24 + 26}{4} = \frac{92}{4} = 23$
    $$

**Answer:** $23$

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