1979 AMC 12 Problem 14

Attempt Problem 14 of the 1979 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 1979 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

In a certain sequence of numbers, the first number is

1,

and, for all

n\ge2,

the product of the first

n

numbers in the sequence is

n^2.

The sum of the third and the fifth numbers in the sequence is

\frac{25}{9}

\frac{31}{15}

\frac{61}{16}

\frac{576}{225}

34

Divide the product of the first

n

terms by the product of the first

n-1

terms

This gives the

n

th term directly as a ratio of two squares

For

n\ge2,

the

n

th term is

a_n=\frac{n^2}{(n-1)^2}.

Therefore

a_3=\frac{9}{4}

and

a_5=\frac{25}{16},

whose sum is

\frac{36}{16}+\frac{25}{16}=\frac{61}{16}.

Therefore, the correct answer is C.

Problem 14 in other years