Difference between revisions of "1952 AHSME Problems/Problem 30"
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== Solution == | == Solution == | ||
Let our first term be <math>a</math> and our common difference be <math>d</math>. Thus, the first few terms of the sequence are <math>a</math>, <math>a + d</math>, <math>a + 2d</math>, ... | Let our first term be <math>a</math> and our common difference be <math>d</math>. Thus, the first few terms of the sequence are <math>a</math>, <math>a + d</math>, <math>a + 2d</math>, ... | ||
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The sum of the first 5 terms is | The sum of the first 5 terms is | ||
<cmath>a + (a + d) + (a + 2d) + ... + (a + 4d) = 5a + 10d</cmath> | <cmath>a + (a + d) + (a + 2d) + ... + (a + 4d) = 5a + 10d</cmath> | ||
Latest revision as of 19:02, 22 December 2015
Problem
When the sum of the first ten terms of an arithmetic progression is four times the sum of the first five terms, the ratio of the first term to the common difference is:
Solution
Let our first term be 
and our common difference be 
. Thus, the first few terms of the sequence are , 
, 
, ...
The sum of the first 5 terms is
![\[a + (a + d) + (a + 2d) + ... + (a + 4d) = 5a + 10d\]](http://latex.artofproblemsolving.com/1/6/7/1670b766d4314b99fda9d82c0c46ea856b5a7043.png)
The sum of the first 10 terms is
![\[a + (a + d) + (a + 2d) + ... + (a + 9d) = 10a + 45d\]](http://latex.artofproblemsolving.com/8/f/6/8f690342bd1afdb18a8b513fa7b3879f9baf2db4.png)
We are told that the latter is four times the former. Hence,
![\[4(5a + 10d) = 10a + 45d\]](http://latex.artofproblemsolving.com/5/b/0/5b0439274d391e3a447d89fb49099ac996da9ae2.png)
![\[20a + 40d = 10a + 45d\]](http://latex.artofproblemsolving.com/4/f/1/4f1279c93c82c0356d6dab7a25f86b8e013b5ce7.png)
![\[10a = 5d\]](http://latex.artofproblemsolving.com/5/0/6/5060381514c302988ca32f84b665180fb63de7de.png)
![\[\frac{a}{d} = \frac{1}{2}\]](http://latex.artofproblemsolving.com/3/f/d/3fd17c69b72edf9671e12fe7a792ff9773715576.png)
Therefore, our answer is 
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Problem 31 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
