Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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...")
 
m (→‎Solution)
 
Line 13: Line 13:
 
Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
−
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>.
+
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 13:10, 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1979_AHSME_Problems/Problem_14&oldid=82212"