# Difference between revisions of "1970 AHSME Problems/Problem 19"

## Problem

The sum of an infinite geometric series with common ratio $r$ such that $|r|<1$ is $15$, and the sum of the squares of the terms of this series is $45$. The first term of the series is

$\textbf{(A) } 12\quad \textbf{(B) } 10\quad \textbf{(C) } 5\quad \textbf{(D) } 3\quad \textbf{(E) 2}$

## Solution

We know that the formula for the sum of an infinite geometric series is $S = \frac{a}{1-r}$.

So we can apply this to the conditions given by the problem.

We have two equations:

\begin{align*} 15 &= \frac{a}{1-r} \\ 45 &= \frac{a^{2}}{1-r^{2}} \end{align*}

We get

\begin{align*} a &= 15 - 15r \\ a^{2} &= 45 - 45r^{2} \\ \\ (15 - 15r)^{2} &= 45 - 45r^{2} \\ 270r^{2} - 450r + 180 &= 0 \\ 3r^{2} - 5r + 2 &= 0 \\ (3r - 2)(r - 1) &= 0 \end{align*}

Since $|r| < 1$, $r = \frac{2}{3}$, so plug this into the equation above and we get $a = 15 - 15r = 15 - 10 = \boxed{\textbf{(C)} \quad 5}$

Solution by $\underline{\textbf{Invoker}}$