# 1979 AHSME Problems/Problem 14

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## Problem 14

In a certain sequence of numbers, the first number is $1$, and, for all $n\ge 2$, 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 $\textbf{(A) }\frac{25}{9}\qquad \textbf{(B) }\frac{31}{15}\qquad \textbf{(C) }\frac{61}{16}\qquad \textbf{(D) }\frac{576}{225}\qquad \textbf{(E) }34$

## Solution

Solution by e_power_pi_times_i

Since the product of the first $n$ numbers in the sequence is $n^2$, the product of the first $n+1$ numbers in the sequence is $(n+1)^2$. Therefore the $n+1$th number in the sequence is $\frac{(n+1)^2}{n^2}$. Therefore the third and the fifth numbers are $\frac{9}{4}$ and $\frac{25}{16}$ respectively. The sum of those numbers is $\frac{36}{16}+\frac{25}{16} = \boxed{\textbf{(C) } \frac{61}{16}}$.

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