2022 AMC 10A Problem 20

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

Answer: E

Concepts:arithmetic sequence geometric sequence Diophantine Equation

Difficulty rating: 2230

Small hint:

Subtract consecutive resulting terms

Big hint:

Use the integer factorization of

b(r-1)^2

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.

Let

t=b(r-1)=br-b,

which is an integer. Then

t^2=28b,

so

t=14u

and

b=7u^2

for some nonzero integer

u.

Because

a=57-b>0,

we must have

u\in\{-2,-1,1,2\}.

Now

r=1+\frac{t}{b}=1+\frac{2}{u}.

The cases

u=-1

and

u=-2

give

r=-1

and

r=0,

respectively, so they cannot produce a positive geometric sequence. If

u=2,

then

b=28, r=2, a=29,

and

d=3-t=-25,

making the third arithmetic term negative.

Therefore

u=1,

so

b = 7, 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.