Difference between revisions of "1962 AHSME Problems/Problem 14"
(Created page with "==Problem== Let <math>s</math> be the limiting sum of the geometric series <math>4- \frac83 + \frac{16}{9} - \dots</math>, as the number of terms increases without bound. Then <m...") |
m (→Solution) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | + | The infinite sum of a geometric series with first term <math>a</math> and common ratio <math>r</math> (<math>-1<r<1</math>) is <math>\frac{a}{1-r}</math>. | |
| + | Now, in this geometric series, <math>a=4</math>, and <math>r=-\frac23</math>. Plugging these into the formula, we get | ||
| + | <math>\frac4{1-(-\frac23)}</math>, which simplifies to <math>\frac{12}5</math>, or <math>\boxed{2.4\textbf{ (B)}}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1962|before=Problem 13|num-a=15}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:16, 3 October 2014
Problem
Let 
be the limiting sum of the geometric series 
, as the number of terms increases without bound. Then equals:
Solution
The infinite sum of a geometric series with first term 
and common ratio 
(
) is 
.
Now, in this geometric series, 
, and 
. Plugging these into the formula, we get
, which simplifies to 
, or 
.
See Also
| 1962 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
