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 "1975 AHSME Problems/Problem 16"

(→‎Solution)
(→‎Solution)
Line 15: Line 15:
  
 
<math>a =</math> The first term in the geometric sequence.
 
<math>a =</math> The first term in the geometric sequence.
 +
 
<math>r =</math> The ratio relating the terms of the geometric sequence.
 
<math>r =</math> The ratio relating the terms of the geometric sequence.
 +
 
<math>n =</math> The nth value of the geometric sequence, starting at 1 and increasing as consecutive integer values.
 
<math>n =</math> The nth value of the geometric sequence, starting at 1 and increasing as consecutive integer values.
  

Revision as of 06:26, 1 July 2024

Problem

If the first term of an infinite geometric series is a positive integer, the common ratio is the reciprocal of a positive integer, and the sum of the series is $3$, then the sum of the first two terms of the series is

$\textbf{(A)}\ \frac{1}{3} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{8}{3} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \frac{9}{2} \qquad$

Solution

Let's establish some ground rules...

$a =$ The first term in the geometric sequence.

$r =$ The ratio relating the terms of the geometric sequence.

$n =$ The nth value of the geometric sequence, starting at 1 and increasing as consecutive integer values.

Using these terms, the sum can be written as: $sum = a/(1-r) = 3$

Let $x =$ The positive integer that is in the reciprocal of the geometric ratio.

This gives: $3 = a/(1-(1/x))$ $3 = ax/(x-1)$

Now through careful inspection we notice that when x = 3 the equation becomes $3 = a(3/2)$, where $a = 2$.

Therefore $r = 1/x = 1/3$. Now we define the sum as $2 * (1/3)^{n-1}$.

Now we simply add the $n = 1$ and $n = 2$ terms.

$sum = 2(1/3)^{(1)-1} + 2(1/3)^{(2)-1} = 2 + 2/3 = 6/3 + 2/3 = 8/3$

This gives $\boxed{\textbf{(C) } 8/3}$.

~PhysicsDolphin

For more information on geometric sequences: https://artofproblemsolving.com/wiki/index.php/Geometric_sequence

See Also

1975 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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=1975_AHSME_Problems/Problem_16&oldid=221390"