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>\ | + | <math>\textbf{(A) } 12\quad |
− | \ | + | \textbf{(B) } 10\quad |
− | \ | + | \textbf{(C) } 5\quad |
− | \ | + | \textbf{(D) } 3\quad |
− | \ | + | \textbf{(E) 2} </math> |
== Solution == | == Solution == | ||
− | <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 == | ||
− | {{AHSME box|year=1970|num-b=18|num-a=20}} | + | {{AHSME 35p box|year=1970|num-b=18|num-a=20}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:14, 20 February 2019
Problem
The sum of an infinite geometric series with common ratio such that is , and the sum of the squares of the terms of this series is . The first term of the series is
Solution
We know that the formula for the sum of an infinite geometric series is .
So we can apply this to the conditions given by the problem.
We have two equations:
We get
Since , , so plug this into the equation above and we get
Solution by
See also
1970 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AHSME Problems and Solutions |
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