2022 AMC 10A Problem 20

P16-20~6 minPrint

Sequences & Series Diophantine Equations Split into exhaustive cases and handle each Set up the equation/formula and compute; no special trick needed

A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are $57$ , $60$ , and $91$ . What is the fourth term of this sequence?

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Key insight

Subtracting neighboring terms twice removes the arithmetic part and leaves b(r - 1) squared = 28. So the ratio r is 2 or 3.

Solution

Step 1: name the sequences. Let the arithmetic sequence be $a,\ a + d,\ a + 2d,\ a + 3d$ and the geometric one $b,\ br,\ br^2,\ br^3$ . Adding term by term:
$a + b = 57, \qquad a + d + br = 60, \qquad a + 2d + br^2 = 91.$

Step 2: subtract neighbors to remove $a$ .
- Second minus first: $d + br - b = 3$ .
- Third minus second: $d + br^2 - br = 31$ .

Step 3: subtract again to remove $d$ . $br^2 - 2br + b = 28$ , which factors as
$b(r - 1)^2 = 28.$

Step 4: which ratios are possible? First, $r$ is a whole number. (If $r$ were a fraction with bottom $q$ in lowest terms, then $br^3$ being whole would need $b$ to be a multiple of $q^3$ . Since $b < 57$ , $q$ could only be $2$ or $3$ , and checking those few values of $b$ shows none of them gives $b(r-1)^2 = 28$ .) Also $r = 1$ would give $0 = 28$ . So $r \ge 2$ , and $(r - 1)^2$ is a perfect square that divides $28$ : it is $1$ or $4$ . So $r = 2$ or $r = 3$ .

Step 5: test both.
- $r = 2$ : $b = 28$ , then $d = 3 - 28 = -25$ and $a = 57 - 28 = 29$ . The arithmetic terms are $29, 4, -21, \ldots$ , and $-21$ is not positive. Rejected.
- $r = 3$ : $b = 28 \div 4 = 7$ , then $d = 3 - (21 - 7) = -11$ and $a = 50$ . Arithmetic: $50, 39, 28, 17$ . Geometric: $7, 21, 63, 189$ . Sums: $57, 60, 91$ , as required.

Step 6: the fourth term. $17 + 189 = 206$ .

The answer is $\boxed{\textbf{(E)}\ 206}$ .

Why this works

Taking differences of neighboring terms removes the starting value of an arithmetic sequence, and taking differences again removes its common difference too. What is left belongs only to the geometric part, and whole-number conditions leave just a couple of ratios to test. Always rebuild both sequences at the end and check every condition, including "positive".

Concise solution

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

Let the arithmetic sequence be $a, a+d, a+2d, a+3d$ and the geometric sequence $b, br, br^2, br^3$ , all positive integers. Then
$a + b = 57, \qquad a + d + br = 60, \qquad a + 2d + br^2 = 91.$
Subtract consecutive equations to eliminate $a$ :
$d + b(r-1) = 3, \qquad d + br(r-1) = 31.$
Subtract again to eliminate $d$ : $b(r-1)^2 = 28$ .

The ratio $r$ is a positive integer greater than $1$ (a fractional ratio $p/q$ would need $q^3 \mid b$ with $b < 57$ , and $q = 2, 3$ fail quickly). So $(r-1)^2$ is a square divisor of $28$ : $1$ or $4$ .

$r = 2$ : $b = 28$ , $d = 3 - 28 = -25$ , $a = 29$ . The arithmetic terms $29, 4, -21, \dots$ are not all positive. Rejected.
$r = 3$ : $b = 7$ , $d = 3 - 14 = -11$ , $a = 50$ . Arithmetic: $50, 39, 28, 17$ ; geometric: $7, 21, 63, 189$ . Check: $57, 60, 91$ . Valid.

The fourth term is $17 + 189 = 206$ .

The answer is $\boxed{\textbf{(E)}\ 206}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)