Difference between revisions of "2004 AMC 12A Problems/Problem 14"

Revision as of 17:21, 11 June 2017

The following problem is from both the 2004 AMC 12A #14 and 2004 AMC 10A #18, so both problems redirect to this page.

Problem

A sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression?

$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$

Solution 1

Let $d$ be the common difference. Then $9$ , $9+d+2=11+d$ , $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , and the third term is $2(-14) + 29 = 1\Rightarrow\boxed{\mathrm{(A)}\ 1}$ .

Solution 2

Let $d$ be the common difference and $r$ be the common ratio. Then the arithmetic sequence is $9$ , $9+d$ , and $9+2d$ . The geometric sequence (when expressed in terms of $d$ ) has the terms $9$ , $11+d$ , and $29+2d$ . Thus, we get the following equations:

$9r=11+d\Rightarrow d=9r-11$

$9r^2=29+2d$

Plugging in the first equation into the second, our equation becomes $9r^2=29+18r-22\Longrightarrow9r^2-18r+7=0$ . By the quadratic formula, $r$ can either be $-\frac{1}{3}$ or $\frac{7}{3}$ . If $r$ is $-\frac{1}{3}$ , the third term (of the geometric sequence) would be $1$ , and if $r$ is $\frac{7}{3}$ , the third term would be $49$ . Clearly the minimum possible value for the third term of the geometric sequence is $\boxed{\mathrm{(A)}\ 1}$ .

Solution 3

Let the three numbers be, in increasing order, $z,y,9$

Hence, we have that $9-y=y-z\implies 9+z=2y$ .

Also, from the second part of information given, we get that

$\frac{9}{y+2}=\frac{y+2}{z+20}\implies 9(z+20)=(y+2)^2\implies y=3(\sqrt{z+20})-2

Plugging back in..$ (Error compiling LaTeX. Unknown error_msg)9+z=6(\sqrt{z+20})-2\implies (9+z)^2=36(z+20)

Simplifying, we get that $z^2-10z-551=0$

Applying the quadratic formula, we get that $z=\frac{10\pm \sqrt{2304}}{2}\implies \frac{10\pm48}{2}

Obviously, in order to minimize the value of$ (Error compiling LaTeX. Unknown error_msg)z $, we have to subtract. Hence,$ z=-19$However, the problem asks for the minimum value of the third term in a geometric progression.

Hence, the answer is$ (Error compiling LaTeX. Unknown error_msg)-19+20=\boxed{1}\implies\boxed{\mathrm{(A)}$

@@ Line 17: / Line 17: @@
 Plugging in the first equation into the second, our equation becomes <math>9r^2=29+18r-22\Longrightarrow9r^2-18r+7=0</math>. By the quadratic formula, <math>r</math> can either be <math>-\frac{1}{3}</math> or <math>\frac{7}{3}</math>.  If <math>r</math> is <math>-\frac{1}{3}</math>, the third term (of the geometric sequence) would be <math>1</math>, and if <math>r</math> is <math>\frac{7}{3}</math>, the third term would be <math>49</math>.  Clearly the minimum possible value for the third term of the geometric sequence is <math>\boxed{\mathrm{(A)}\ 1}</math>.
+==Solution 3==
+Let the three numbers be, in increasing order, <math>z,y,9</math>
+Hence, we have that <math>9-y=y-z\implies 9+z=2y</math>.
+Also, from the second part of information given, we get that
+<math>\frac{9}{y+2}=\frac{y+2}{z+20}\implies 9(z+20)=(y+2)^2\implies y=3(\sqrt{z+20})-2
+Plugging back in..
+</math>9+z=6(\sqrt{z+20})-2\implies (9+z)^2=36(z+20)
+Simplifying, we get that <math>z^2-10z-551=0</math>
+Applying the quadratic formula, we get that <math>z=\frac{10\pm \sqrt{2304}}{2}\implies \frac{10\pm48}{2}
+Obviously, in order to minimize the value of </math>z<math>, we have to subtract. Hence, </math>z=-19<math>
+However, the problem asks for the minimum value of the third term in a geometric progression.
+Hence, the answer is </math>-19+20=\boxed{1}\implies\boxed{\mathrm{(A)}$
 == See also ==

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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
All AMC 12 Problems and Solutions

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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
All AMC 10 Problems and Solutions

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