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2014 AMC 10A Problem 10

2014 AMC 10A Problem 10AMC 10 Step by Step

Name: Date: September 13, 2026

Ratios, Percents & Averages · difficulty 1 of 5 · about 1 min · position P1-10

Five positive consecutive integers starting with $a$ have average $b$ . What is the average of $5$ consecutive integers that start with $b$ ?

A)
$a+3$
B)
$a+4$
C)
$a+5$
D)
$a+6$
E)
$a+7$

Answer

Key insight

The average of five consecutive integers is the middle one, so b = a + 2 and the new average is b + 2 = a + 4.

Solution

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

Why this works

For any arithmetic progression with an odd number of terms, the mean equals the median, the middle term, because terms pair off symmetrically around it. Applying that fact twice avoids any summation. With $a = 1$ : the first list is $1$ to $5$ with average $3$ , and $3$ to $7$ has average $5 = a + 4$ , a quick confirmation.

The trap

Adding the five numbers and dividing without noticing the shortcut, then slipping and getting a + 3 or a + 5.

Common mistakes

Adding the five numbers and dividing without noticing the shortcut, then slipping and getting a + 3 or a + 5.
Answering $b + 2$ and forgetting to express the result in terms of $a$ .

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer