AoPSWiki
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
Personal tools

2009 AMC 12B Problems/Problem 12

From AoPSWiki

Problem

The fifth and eighth terms of a geometric sequence of real numbers are 7! and 8! respectively. What is the first term?


\mathrm{(A)}\ 60\qquad\mathrm{(B)}\ 75\qquad\mathrm{(C)}\ 120\qquad\mathrm{(D)}\ 225\qquad\mathrm{(E)}\ 315

Solution

Let the nth term of the series be ar^{n-1}. Because \frac {8!}{7!} = \frac {ar^7}{ar^4} = r^3 = 8, it follows that r = 2 and the first term is a = \frac {7!}{r^4} = \frac {7!}{16} = \boxed{315}. The answer is \mathrm{(E)}.

See also

2009 AMC 12B (Problems • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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
Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2009_AMC_12B_Problems/Problem_12"
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us