Difference between revisions of "1964 AHSME Problems/Problem 6"
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− | 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>. | ||
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<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> | + | <math>x^2+5x+4 = 0</math> |
+ | <math>(x + 4)(x+1) = 0</math> | ||
− | <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 | + | 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> |
− | + | ==See Also== | |
+ | {{AHSME 40p box|year=1964|num-b=5|num-a=7}} | ||
− | + | [[Category:Introductory Algebra Problems]] | |
− | + | {{MAA Notice}} |
Latest revision as of 03:12, 23 July 2019
Problem
If are in geometric progression, the fourth term is:
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:
.
Solving it, we get:
or
If , the sequence has a 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 , which is .
If , we plug into to find the sequence starts as . The common ratio is . The next term is , which is option
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.