2004 AMC 10A Problem 18

Question

A sequence of three real numbers forms an arithmetic progression with a first term of

9.

If

2

is added to the second term and

20

is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?

1

4

36

49

81

Answer: A

Concepts:arithmetic sequence geometric sequence quadratic

Difficulty rating: 1630

Small hint:

Write the progression as

9,

9 + d,

9 + 2d,

so the geometric one is

9,

11 + d,

29 + 2d

Big hint:

Use

(11 + d)^2 = 9(29 + 2d)

and take the value of

d

giving the smaller third term

Solution:

The arithmetic progression is

9,

9 + d,

9 + 2d,

so the geometric progression is

9,

11 + d,

29 + 2d.

The geometric condition gives

(11 + d)^2 = 9(29 + 2d),

which simplifies to

d^2 + 4d - 140 = 0,

so

d = 10

or

d = -14.

The third terms

29 + 2d

are

49

and

1.

The smallest is

1.

Thus, the correct answer is A.