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

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The sum of an infinite geometric series with common ratio <math>r</math> such that <math>|r|<1</math> is <math>15</math>, and the sum of the squares of the terms of this series is <math>45</math>. The first term of the series is
 
The sum of an infinite geometric series with common ratio <math>r</math> such that <math>|r|<1</math> is <math>15</math>, and the sum of the squares of the terms of this series is <math>45</math>. The first term of the series is
  
−
<math>\text{(A) } 12\quad
+
<math>\textbf{(A) } 12\quad
−
\text{(B) } 10\quad
+
\textbf{(B) } 10\quad
−
\text{(C) } 5\quad
+
\textbf{(C) } 5\quad
−
\text{(D) } 3\quad
+
\textbf{(D) } 3\quad
−
\text{(E) 2} </math>
+
\textbf{(E) 2} </math>
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{C}</math>
+
We know that the formula for the sum of an infinite geometric series is <math>S = \frac{a}{1-r}</math>.
 +
 
 +
So we can apply this to the conditions given by the problem.
 +
 
 +
We have two equations:
 +
 
 +
<cmath>
 +
\begin{align*}
 +
15 &= \frac{a}{1-r} \\
 +
45 &= \frac{a^{2}}{1-r^{2}}
 +
\end{align*}
 +
</cmath>
 +
 
 +
We get
 +
 
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 
 +
Since <math>|r| < 1</math>, <math>r = \frac{2}{3}</math>, so plug this into the equation above and we get <math>a = 15 - 15r = 15 - 10 = \boxed{\textbf{(C)} \quad 5}</math>
 +
 
 +
Solution by <math>\underline{\textbf{Invoker}}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 20:14, 20 February 2019

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 $\underline{\textbf{Invoker}}$

See also

1970 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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