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# 2002 AIME II Problems/Problem 11

## Problem

Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$, and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$, where $m$, $n$, and $p$ are positive integers and $m$ is not divisible by the square of any prime. Find $100m+10n+p$.

## Solution 1

Let the second term of each series be $x$. Then, the common ratio is $\frac{1}{8x}$, and the first term is $8x^2$.

So, the sum is $\frac{8x^2}{1-\frac{1}{8x}}=1$. Thus, $64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}$.

The only solution in the appropriate form is $x = \frac{\sqrt{5}-1}{8}$. Therefore, $100m+10n+p = \boxed{518}$.

## Solution 2

Let the two sequences be $a, ar, ar^2 ... \text{ }an^2$ and $x, xy, xy^2 ... \text{ }xy^n$. We know for a fact that $ar = xy$. We also know that the sum of the first sequence = $\frac{a}{1-r} = 1$, and the sum of the second sequence = $\frac{x}{1-y} = 1$. Therefore we have $$a+r = 1$$$$x+y = 1$$$$ar=xy$$ We can then replace $r = \frac{xy}{a}$ and $y = \frac{ar}{x}$. We plug them into the two equations $a+r = 1$ and $x+y = 1$. We then get $$x^2 + ar = x$$$$a^2 + xy = a$$We subtract these equations, getting $$x^2 - a^2 + ar - xy = x-a$$Remember $$ar=xy$$, so $$(x-a)(x+a-1) = 0$$Then considering cases, we have either $x=a$ or $y=a$. This suggests that the second sequence is in the form $r, ra, ra^2...$, while the first sequence is in the form $a, ar, ar^2...$ Now we have that $ar^2 = \frac18$ and we also have that $a+r = 1$. We can solve for $r$ and the only appropriate value for $r$ is $\frac{1+\sqrt{5}}{4}$. All we want is the second term, which is $ar = \frac{ar^2}{r} = \frac{\frac18}{\frac{1+\sqrt{5}}{4}} = \frac{\sqrt{5} - 1}{8}$ solution by jj_ca888