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 "1964 AHSME Problems/Problem 6"

(Created page with "Since we know the sequence is a geometric sequence, we can set up an equation: <math>\frac{2x+2}{x} = \frac{3x+3}{2x+2}</math>. Solving it, we get: <math>\frac{2x+2}{x} = \...")
 
 
Line 1: Line 1:
−
Since we know the sequence is a geometric sequence, we can set up an equation:
+
== Problem ==
 +
 
 +
If <math>x, 2x+2, 3x+3, \dots</math> are in geometric progression, the fourth term is:
 +
 
 +
<math>\textbf{(A)}\ -27 \qquad
 +
\textbf{(B)}\ -13\frac{1}{2} \qquad
 +
\textbf{(C)}\ 12\qquad
 +
\textbf{(D)}\ 13\frac{1}{2}\qquad
 +
\textbf{(E)}\ 27 </math> 
 +
 
 +
 
 +
==Solution==
 +
 
 +
Since we know the sequence is a geometric sequence, the ratio of consecutive terms is always the same number.  Thus, we can set up an equation:
  
 
<math>\frac{2x+2}{x} = \frac{3x+3}{2x+2}</math>.
 
<math>\frac{2x+2}{x} = \frac{3x+3}{2x+2}</math>.
Line 7: Line 20:
 
<math>\frac{2x+2}{x} = \frac{3x+3}{2x+2}</math>
 
<math>\frac{2x+2}{x} = \frac{3x+3}{2x+2}</math>
  
 +
<math>(2x+2)(2x+2) = (3x+3)(x)</math>
 +
 +
<math>4x^2+8x+4 = 3x^2+3x</math>
  
−
<math>(2x+2)(2x+2) = (3x+3)(x)</math>
+
<math>x^2+5x+4 = 0</math>
  
 +
<math>(x + 4)(x+1) = 0</math>
  
−
<math>4x^2+8x+4 = 3x^2+3</math>
+
<math>x = -4</math> or <math>x = -1</math>
  
 +
If <math>x=-1</math>, the sequence has a <math>0</math> as the second term, which is not allowed in a geometric sequence, so it is an extraneous solution that came about because we cross-multiplied by <math>(2x + 2)</math>, which is <math>0</math>.
  
−
<math>x^2+8x+1 = 0</math>.
+
If <math>x=-4</math>, we plug into <math>x, 2x +2, 3x + 3</math> to find the sequence starts as <math>-4, -6, -9</math>.  The common ratio is <math>\frac{-6}{-4} = \frac{3}{2}</math>. The next term is <math>\frac{3}{2} \cdot -9 = \frac{-27}{2} = -13\frac{1}{2}</math>, which is option <math>\textbf{(B)}</math>
  
−
Applying the quadratic formula, we get
+
==See Also==
 +
{{AHSME 40p box|year=1964|num-b=5|num-a=7}}
  
−
<math>\frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-8\pm \sqrt{8^2-4(1)(1)}}{2(1)} = \frac{-8\pm \sqrt{64-4}}{2} = \frac{-8\pm 2\sqrt{15}}{2} = -4\pm \sqrt{15}</math>.
+
[[Category:Introductory Algebra Problems]]
  
−
From there, you can plug <math>x</math> in to find out the answer.
+
{{MAA Notice}}

Latest revision as of 04:12, 23 July 2019

Problem

If $x, 2x+2, 3x+3, \dots$ are in geometric progression, the fourth term is:

$\textbf{(A)}\ -27 \qquad \textbf{(B)}\ -13\frac{1}{2} \qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 13\frac{1}{2}\qquad \textbf{(E)}\ 27$


Solution

Since we know the sequence is a geometric sequence, the ratio of consecutive terms is always the same number. Thus, we can set up an equation:

$\frac{2x+2}{x} = \frac{3x+3}{2x+2}$.

Solving it, we get:

$\frac{2x+2}{x} = \frac{3x+3}{2x+2}$

$(2x+2)(2x+2) = (3x+3)(x)$

$4x^2+8x+4 = 3x^2+3x$

$x^2+5x+4 = 0$

$(x + 4)(x+1) = 0$

$x = -4$ or $x = -1$

If $x=-1$, the sequence has a $0$ as the second term, which is not allowed in a geometric sequence, so it is an extraneous solution that came about because we cross-multiplied by $(2x + 2)$, which is $0$.

If $x=-4$, we plug into $x, 2x +2, 3x + 3$ to find the sequence starts as $-4, -6, -9$. The common ratio is $\frac{-6}{-4} = \frac{3}{2}$. The next term is $\frac{3}{2} \cdot -9 = \frac{-27}{2} = -13\frac{1}{2}$, which is option $\textbf{(B)}$

See Also

1964 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
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=1964_AHSME_Problems/Problem_6&oldid=107814"