2004 AMC 12A Problems/Problem 14

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


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:

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


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..

$9+z=6(\sqrt{z+20})-4\implies (9+z)^2=36(z+20)$

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 $z$, 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 $-19+20=\boxed{1} \implies \boxed{A}$

Solution 4

Let the arithmetic sequence be $9,9+x,9+2x$ and let the geometric sequence be $9,11+x,29+2x$. Now, we just try all the solutions. If the last term is $1$, then $x=-14$. This gives the geometric sequence $9,-3,1$ which indeed works. The answer is $\boxed{1} \implies \boxed{A}$

Solution by franzliszt

Solution 5

The terms of the arithmetic progression are 9, $9+d$, and $9+2d$ for some real number $d$. The terms of the geometric progression are 9, $11+d$, and $29+2d$. Therefore$(11+d)^{2} = 9(29+2d) \quad\text{so}\quad d^{2}+4d-140 = 0.$Thus $d=10$ or $d=-14$. The corresponding geometric progressions are $9, 21, 49$ and $9, -3, 1,$ so the smallest possible value for the third term of the geometric progression is $\boxed{1} \implies \boxed{A}$.

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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