1977 AMC 12 Problem 13

Attempt Problem 13 of the 1977 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1977 AMC 12 solutions, or check the answer key.

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13.

If a1,a_1, a2,a_2, a3,a_3, \ldots is a sequence of positive numbers such that an+2=anan+1a_{n+2}=a_na_{n+1} for all positive integers n,n, then the sequence a1,a_1, a2,a_2, a3,a_3, \ldots is a geometric progression

for all positive values of a1a_1 and a2a_2

if and only if a1=a2a_1=a_2

if and only if a1=1a_1=1

if and only if a2=1a_2=1

if and only if a1=a2=1a_1=a_2=1

Answer: E
Concepts:recursiongeometric sequenceratio and proportion
Difficulty rating: 1860
Small Hint:

Write out a3,a_3, a4,a_4, and a5a_5 in terms of a1,a_1, a2a_2

Big Hint:

Equate the first three successive ratios and use positivity

Solution:

The next terms are a3=a1a2,a_3=a_1a_2, a4=a1a22,a_4=a_1a_2^2, and a5=a12a23.a_5=a_1^2a_2^3. If the sequence is geometric, then a2a1=a3a2=a1,a4a3=a2=a1. \begin{aligned} \frac{a_2}{a_1}&=\frac{a_3}{a_2}=a_1,\\ \frac{a_4}{a_3}&=a_2=a_1. \end{aligned} Thus a2=a12a_2=a_1^2 and a2=a1.a_2=a_1. Positivity forces a1=a2=1.a_1=a_2=1. Conversely, these initial values produce the constant geometric sequence 1,1,1,.1,1,1,\ldots.

Therefore, the correct answer is E.

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