Difference between revisions of "2022 AMC 10B Problems/Problem 15"
(→Solution) |
Numerophile (talk | contribs) (→Solution) |
||
| Line 5: | Line 5: | ||
<math>\textbf{(A) } 340 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 380 \qquad \textbf{(D) } 400 \qquad \textbf{(E) } 420</math> | <math>\textbf{(A) } 340 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 380 \qquad \textbf{(D) } 400 \qquad \textbf{(E) } 420</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Suppose that the first number of the arithmetic sequence is <math>a</math>. We will try to compute the value of <math>S_{n}</math>. 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 <math>a + n - 1</math>. Thus, the value of <math>S_{n}</math> is <math>n(a + n - 1) = n^2 + n(a - 1)</math>. Then, <cmath>\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)}.</cmath> Of course, for this value to be constant, <math>6n(a-1)</math> must be <math>0</math> for all values of <math>n</math>, and thus <math>a = 1</math>. Finally, the value of <math>S_{20}</math> is <math>20^2 = \fbox{D. 400}</math> | Suppose that the first number of the arithmetic sequence is <math>a</math>. We will try to compute the value of <math>S_{n}</math>. 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 <math>a + n - 1</math>. Thus, the value of <math>S_{n}</math> is <math>n(a + n - 1) = n^2 + n(a - 1)</math>. Then, <cmath>\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)}.</cmath> Of course, for this value to be constant, <math>6n(a-1)</math> must be <math>0</math> for all values of <math>n</math>, and thus <math>a = 1</math>. Finally, the value of <math>S_{20}</math> is <math>20^2 = \fbox{D. 400}</math> | ||
~mathboy100 | ~mathboy100 | ||
| + | |||
| + | ==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> | ||
| + | |||
| + | ~numerophile | ||
| + | |||
== See Also == | == See Also == | ||
{{AMC10 box|year=2022|ab=B|num-b=14|num-a=16}} | {{AMC10 box|year=2022|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:40, 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. 
