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 "1974 AHSME Problems/Problem 21"

(Created page with "==Problem== In a geometric series of positive terms the difference between the fifth and fourth terms is <math> 576 </math>, and the difference between the second and first terms...")
 
 
(One intermediate revision by one other user not shown)
Line 11: Line 11:
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1974|num-b=20|num-a=22}}
 
{{AHSME box|year=1974|num-b=20|num-a=22}}
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:43, 5 July 2013

Problem

In a geometric series of positive terms the difference between the fifth and fourth terms is $576$, and the difference between the second and first terms is $9$. What is the sum of the first five terms of this series?

$\mathrm{(A)\ } 1061 \qquad \mathrm{(B) \ }1023 \qquad \mathrm{(C) \ } 1024 \qquad \mathrm{(D) \ } 768 \qquad \mathrm{(E) \ }\text{none of these}$

Solution

Let the first term be $a$ and the common ratio be $r$. Therefore, the second term is $ar$, the fourth term is $ar^3$, and the fifth term is $ar^4$. We're given that $ar^4-ar^3=576\implies ar^3(r-1)=576$ and $ar-a=9\implies a(r-1)=9$. Dividing this first equation by this second one, we get $r^3=\frac{576}{9}=64\implies r=4$. Therefore, $a(4-1)=9$, so $a=3$.

Therefore, the first five terms of this series are $3, 12, 48, 192, 768$, and their sum is $1023, \boxed{\text{B}}$.

See Also

1974 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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=1974_AHSME_Problems/Problem_21&oldid=56273"