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 term of this sequence?

$190$

$194$

$198$

$202$

$206$

Solution:

Let the arithmetic sequence be $a, a + d, a + 2d, a + 3d$ and the geometric sequence be $b, br, br^2, br^3.$

Then $a + b = 57 \tag*{(1)},$ $a + d + br = 60 \tag*{(2)},$ and $a + 2d + br^2 = 91 \tag*{(3)}.$

Subtracting $(1)$ from $(2)$ and $(2)$ from $(3),$ we get $d + b(r - 1) = 3$ and $d + br(r - 1) = 31.$

Subtracting these, we get $b(r - 1)^2 = 28.$

Since every variable is an integer, we get that $b = 28$ or $b = 7.$

If $b = 28,$ then $r = 2,$ $a = 29,$ and $d = -25.$ This forces the arithmetic sequence to be $29, 4, -21, -46,$ which is a contradiction.

Therefore, $b = 7.$ Then $r = 3,$ $a = 50,$ and $d = -11.$ The arithmetic sequence is $50,39,28,17,$ and the geometric sequence is $7,21,63,189.$

The desired answer is $17 + 189 = 206.$

Thus, E is the correct answer.

2022 AMC 10A Problem 20

Solution:

Problem 20 in Other Years