2003 AIME I Problem 8

Question

In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by

30.

Find the sum of the four terms.

Answer

Answer: 129

Concepts:arithmetic sequence geometric sequence casework

Difficulty rating: 2210

Small hint:

Write the terms as

a,

a + d,

a + 2d,

and

a + 30

Big hint:

The geometric condition becomes

3a(10 - d) = 2d(2d - 15),

so

10 - d

and

2d - 15

must have the same sign

Solution:

Write the terms as

a,

a + d,

a + 2d,

and

a + 30,

where

a

and

d

are positive integers. The geometric condition on the last three terms says

(a + 30)(a + d) = (a + 2d)^2.

Expanding both sides and simplifying,

30a + 30d = 3ad + 4d^2,

that is

3a(10 - d) = 2d(2d - 15).

Since

a, d \gt 0,

the factors

10 - d

and

2d - 15

must have the same sign, forcing

7.5 \lt d \lt 10,

so

d = 8

or

d = 9.

For

d = 8,

we get

6a = 16,

which has no integer solution. For

d = 9,

we get

3a = 54,

so

a = 18.

The sequence is

18, 27, 36, 48

(indeed

27, 36, 48

has ratio

\frac{4}{3}

), and the sum is

18 + 27 + 36 + 48 = 129.