Difference between revisions of "2022 AMC 10B Problems/Problem 15"
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==Solution 2 (Quick Insight)== | ==Solution 2 (Quick Insight)== | ||
| − | Recall that the sum of the first <math>n</math> odd numbers is <math>n^2</math>. <math>\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9</math>. Thus <math>S_n = 20^2 = \fbox{D. 400}</math> | + | Recall that the sum of the first <math>n</math> odd numbers is <math>n^2</math>. |
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| + | <math>\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9</math>. Thus <math>S_n = 20^2 = \fbox{D. 400}</math> | ||
~numerophile | ~numerophile | ||
Revision as of 21:28, 18 November 2022
Problem
Let 
be the sum of the first 
term of an arithmetic sequence that has a common difference of 
. The quotient 
does not depend on . What is 
?
Solution 1
Suppose that the first number of the arithmetic sequence is 
. We will try to compute the value of 
. First, note that the sum of an arithmetic sequence is equal to the number of terms multiplied by the median of the sequence. The median of this sequence is equal to 
. Thus, the value of is 
. Then, ![\[\frac{S_{3n}}{S_{n}} = \frac{9n^2 + 3n(a - 1)}{n^2 + n(a - 1)} = 9 - \frac{6n(a-1)}{n^2 + n(a-1)}.\]](http://latex.artofproblemsolving.com/d/5/d/d5d476118c91fd24b5e4fb0cf2a9267409654c10.png)
Of course, for this value to be constant, 
must be 
for all values of , and thus 
. Finally, the value of is 
~mathboy100
Solution 2 (Quick Insight)
Recall that the sum of the first odd numbers is 
.
. Thus 
~numerophile
See Also
| 2022 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
