Difference between revisions of "2022 AMC 10B Problems/Problem 15"
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==Solution 2== | ==Solution 2== | ||
− | + | Let's say that our sequence is <cmath>a, a+2, a+4, a+6, a+8, a+10, \ldots.</cmath> | |
+ | Then, since the value of n doesn't matter in the quotient <math>\frac{S_{3n}}{S_n}</math>, we can say that | ||
+ | <cmath>\frac{S_{3}}{S_1} = \frac{S_{6}}{S_2}.</cmath> | ||
+ | Simplifying, we get <math>\frac{3a+6}{a}=\frac{6a+30}{2a+2}</math>, from which <cmath>\frac{3a+6}{a}=\frac{3a+15}{a+1}.</cmath> <cmath>3a^2+9a+6=3a^2+15a</cmath> <cmath>6a=6</cmath> | ||
+ | Solving for <math>a</math>, we get that <math>a=1</math>. | ||
− | the | + | Since the sum of the first <math>n</math> odd numbers is <math>n^2</math>, <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>. |
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==Solution 3 (Quick Insight)== | ==Solution 3 (Quick Insight)== |
Revision as of 15:52, 6 June 2024
Contents
[hide]Problem
Let be the sum of the first
terms 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,
Of course, for this value to be constant,
must be
for all values of
, and thus
. Finally, we have
.
~mathboy100
Solution 2
Let's say that our sequence is
Then, since the value of n doesn't matter in the quotient
, we can say that
Simplifying, we get
, from which
Solving for
, we get that
.
Since the sum of the first odd numbers is
,
.
Solution 3 (Quick Insight)
Recall that the sum of the first odd numbers is
.
Since , we have
.
~numerophile
Video Solution (🚀 Solved in 4 min 🚀)
~Education, the Study of Everything
Video Solution by Interstigation
Video Solution by paixiao
https://www.youtube.com/watch?v=4bzuoKi2Tes
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 |
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