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Difference between revisions of "1979 AHSME Problems/Problem 14"

(Created page with "== Problem 14 == In a certain sequence of numbers, the first number is <math>1</math>, and, for all <math>n\ge 2</math>, the product of the first <math>n</math> numbers in...")
 
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Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
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Since the product of the first <math>n</math> numbers in the sequence is <math>n^2</math>, the product of the first <math>n+1</math> numbers in the sequence is <math>(n+1)^2</math>. Therefore the <math>n+1</math> number in the sequence is <math>\frac{(n+1)^2}{n^2}</math>. Therefore the third and the fifth numbers are <math>\frac{9}{4}</math> and <math>\frac{25}{16}</math> respectively. The sum of those numbers is <math>\frac{36}{16}+\frac{25}{16} = \boxed{\textbf{(C) } \frac{61}{16}}</math>.
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Since the product of the first <math>n</math> numbers in the sequence is <math>n^2</math>, the product of the first <math>n+1</math> numbers in the sequence is <math>(n+1)^2</math>. Therefore the <math>n+1</math>th number in the sequence is <math>\frac{(n+1)^2}{n^2}</math>. Therefore the third and the fifth numbers are <math>\frac{9}{4}</math> and <math>\frac{25}{16}</math> respectively. The sum of those numbers is <math>\frac{36}{16}+\frac{25}{16} = \boxed{\textbf{(C) } \frac{61}{16}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 12:10, 6 January 2017

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 $\frac{36}{16}+\frac{25}{16} = \boxed{\textbf{(C) } \frac{61}{16}}$.

See also

1979 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AHSME Problems and Solutions

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