2022 AMC 10A, Problem 20, Hints and Solution

Motivation

To find the last term in a sequence, each term formed by adding similar indexed term from an AP and a GP.

Question

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 of this sequence?

Hint 1

Use the standard forms of the terms of the progressions to obtain a system of equations.

Hint 2

Try to reduce the number of variables from the system by subtracting two subsequent equations at a time. 

Let $a,ar,ar^2,ar^3$, be the first three terms of the geometric progression, and $b,b+d,b+2d,b+3d$ be the corresponding terms of the arithmetic progression.

We are given, that

$$a+b=57$$

$$ar+b+d=60$$

$$ar^2+b+2d=91.$$

Final Solution

These are 3 non-linear equations in 4 variables, so we can't directly conclude anything. Notice that if we subtract the first two equations we get, discarding $b$ $$3=a(r-1)+d$$ and similarly 

 $$31=ar(r-1)+d.$$

Each of these equations contain, the same variable. So subtracting again, we get

$$28=ar^2-2ar+a=a(r-1)^2.$$

Now since we're dealing with sequences of positive integers, then we can only equate $(r-1)^2$ to either $4$ or $1$.

Then we can conclude that either $a=28$ and $r=2$ or $a=7$ and $r=3$.

If $a=28$, then we get $b=57-28=29$ and $d=-25$. But that makes the arithmetic progression $29,4,-21,-46$, which is a contradiction since the sequence is of positive integers. With $a=7$, $b=50$, and $d=-11$ we get following progressions $50,39,28,17$ and $7,21,63,189$.

The desired number is then

$$17+189=206$$