Difference between revisions of "2022 AMC 10B Problems/Problem 15"
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==Problem== | ==Problem== | ||
− | Let <math>S_n</math> be the sum of the first <math>n</math> | + | Let <math>S_n</math> be the sum of the first <math>n</math> terms of an arithmetic sequence that has a common difference of <math>2</math>. The quotient <math>\frac{S_{3n}}{S_n}</math> does not depend on <math>n</math>. What is <math>S_{20}</math>? |
<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> | ||
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==Solution 2== | ==Solution 2== | ||
− | + | We'll start with the ratio that term 3 ÷ term 1 = term 6 ÷ term 2 | |
− | |||
− | |||
− | |||
− | |||
− | + | the sequence goes like: a, a+2, a+4, a+6, a+8, a+10... | |
+ | |||
+ | term 3 ÷ term 1 = a+4 ÷ a | ||
+ | |||
+ | term 6 ÷ term 2 = a+10 ÷ a+2 | ||
+ | |||
+ | a+4 ÷ a = a+10 ÷ a+2 | ||
+ | |||
+ | (a+4)(a+2) = (a)(a+10) | ||
+ | |||
+ | a^2+6a+8 = a^2+10a | ||
+ | |||
+ | 6a+8=10a | ||
+ | |||
+ | 8=4a | ||
+ | |||
+ | 2=a | ||
+ | |||
+ | |||
+ | the sequence is updated to 2,4,6,8,10,12...40 | ||
+ | |||
+ | or 2(1+2+3+4+5+6...+20) | ||
+ | |||
+ | which is also 2(20 x 21)/2 or 20 x 21 and that is 420. | ||
==Solution 3 (Quick Insight)== | ==Solution 3 (Quick Insight)== | ||
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~numerophile | ~numerophile | ||
− | ==Video Solution | + | ==Video Solution (🚀 Solved in 4 min 🚀)== |
https://youtu.be/7ztNpblm2TY | https://youtu.be/7ztNpblm2TY | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/qkyRBpQHbOA | ||
+ | ==Video Solution by paixiao== | ||
+ | https://www.youtube.com/watch?v=4bzuoKi2Tes | ||
== 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 13:34, 2 April 2024
Contents
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
We'll start with the ratio that term 3 ÷ term 1 = term 6 ÷ term 2
the sequence goes like: a, a+2, a+4, a+6, a+8, a+10...
term 3 ÷ term 1 = a+4 ÷ a
term 6 ÷ term 2 = a+10 ÷ a+2
a+4 ÷ a = a+10 ÷ a+2
(a+4)(a+2) = (a)(a+10)
a^2+6a+8 = a^2+10a
6a+8=10a
8=4a
2=a
the sequence is updated to 2,4,6,8,10,12...40
or 2(1+2+3+4+5+6...+20)
which is also 2(20 x 21)/2 or 20 x 21 and that is 420.
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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