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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
