2026 AMC 10A Problem 5

P1-10~2 minPrint

Sequences & Series Statistics & Data Set up the equation/formula and compute; no special trick needed Compute small cases, spot the pattern, generalize

Let $a_1,a_2,a_3,\dots$ be a sequence with $a_1=94, a_2=190,$ and $a_n$ equal to the arithmetic mean of the previous $n-1$ terms for $n\geq 3$ . What is the value of $a_{10}$ ?

Show solution without answering

Key insight

Adding a new number equal to the current average leaves the average unchanged. So once the third term is 142, every later term is 142 too.

Solution

Each new term is the average of all the terms before it. Compute the first couple of new terms and watch what happens.

1. Third term. It is the average of the first two:
$a_3 = \frac{94 + 190}{2} = \frac{284}{2} = 142.$

2. Fourth term. It is the average of the first three:
$a_4 = \frac{94 + 190 + 142}{3} = \frac{426}{3} = 142.$
The same value again.

See why it keeps happening. The first three terms average $142$ . When you add a new number that equals the average, the average does not move: the new number is neither above nor below it, so it does not pull the average either way. In totals: the first three terms add to $3 \times 142$ ; adding another $142$ makes the total $4 \times 142$ for four terms, and the average is still $142$ .

Keep going. So $a_4 = 142$ leaves the average of the first four terms at $142$ , which makes $a_5 = 142$ , which leaves the average of the first five at $142$ , and so on. Every term from $a_3$ onward is $142$ . In particular $a_{10} = 142$ .

The answer is $\boxed{\textbf{(C)}\ 142}$ .

Why this works

An average is a balance point. Putting a new number exactly at the balance point changes nothing, so a sequence built by "next term = average of everything so far" stops changing after the first new term. When a sequence rule looks like it needs many steps, compute two or three terms first; often the pattern settles down right away.

Alternative approach

Track totals instead of averages. Let $S_n$ be the total of the first $n$ terms. Then $S_2 = 284$ and $a_3 = 284 \div 2 = 142$ , so $S_3 = 426$ . Next $a_4 = 426 \div 3 = 142$ , so $S_4 = 568$ , and $a_5 = 568 \div 4 = 142$ . Each step adds $142$ to the total and one more term to the count, so every total is $142$ times its count, and each new term stays $142$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

$a_3 = \tfrac{94 + 190}{2} = 142$ . Once the first $k$ terms have average $142$ , the next term is $142$ , and appending it gives $k + 1$ terms with total $142(k+1)$ , so the average is still $142$ . Hence $a_n = 142$ for all $n \ge 3$ , and $a_{10} = 142$ .

The answer is $\boxed{\textbf{(C)}\ 142}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Prealgebra, Chapter 13: Data and Statistics
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)