2014 AMC 10A Problem 10

Try a small case first. If $a = 1$, the five numbers are $1, 2, 3, 4, 5$. Their sum is $15$ and their average is $3$, which is the middle number. That always happens: the numbers are spread evenly around the middle one, so the ones below it and above it balance out.

1. The first list is $a, a+1, a+2, a+3, a+4$. Its middle number is $a + 2$, so $b = a + 2$.
2. The second list starts at $b$: $b, b+1, b+2, b+3, b+4$. Its average is its middle number, $b + 2$.
3. Put in $b = a + 2$: the average is $(a + 2) + 2 = a + 4$.

Check with $a = 1$: then $b = 3$, and the list $3, 4, 5, 6, 7$ has average $5$, which is $1 + 4$.

The answer is $\boxed{\textbf{(B)}\ a+4}$.

When numbers go up by the same step and there is an odd number of them, the average is the middle one, because each number below the middle is matched by one the same distance above it. Using that fact twice makes the problem a two-line calculation. Checking with a small value like $a = 1$ is a quick way to be sure.

The five integers $a, a+1, a+2, a+3, a+4$ are evenly spaced, so their average is the middle one: $b = a + 2$.

The five consecutive integers starting with $b$ are $b, b+1, b+2, b+3, b+4$, whose average is likewise the middle term $b + 2$. Substituting $b = a + 2$,
$$b + 2 = (a + 2) + 2 = a + 4.$$

The answer is $\boxed{\textbf{(B)}\ a+4}$.