# Difference between revisions of "2002 AMC 12B Problems/Problem 9"

## Problem

If $a,b,c,d$ are positive real numbers such that $a,b,c,d$ form an increasing arithmetic sequence and $a,b,d$ form a geometric sequence, then $\frac ad$ is

$\mathrm{(A)}\ \frac 1{12} \qquad\mathrm{(B)}\ \frac 16 \qquad\mathrm{(C)}\ \frac 14 \qquad\mathrm{(D)}\ \frac 13 \qquad\mathrm{(E)}\ \frac 12$

## Solution

### Solution 1

We can let a=1, b=2, c=3, and d=4. $\frac{a}{d}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$

### Solution 2

As $a, b, d$ is a geometric sequence, let $b=ka$ and $d=k^2a$ for some $k>0$.

Now, $a, b, c, d$ is an arithmetic sequence. Its difference is $b-a=(k-1)a$. Thus $d=a + 3(k-1)a = (3k-2)a$.

Comparing the two expressions for $d$ we get $k^2=3k-2$. The positive solution is $k=2$, and $\frac{a}{d}=\frac{a}{k^2a}=\frac{1}{k^2}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$.

### Solution 3

Letting $n$ be the common difference of the arithmetic progression, we have $b = a + n$, $c = a + 2n$, $d = a + 3n$. We are given that $b / a$ = $d / b$, or $$\frac{a + n}{a} = \frac{a + 3n}{a + n}.$$ Cross-multiplying, we get $$a^2 + 2an + n^2 = a^2 + 3an$$ $$n^2 = an$$ $$n = a$$ So $\frac{a}{d} = \frac{a}{a + 3n} = \frac{a}{4a} = \boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$.