Difference between revisions of "2022 AMC 10B Problems/Problem 15"
(→Video Solution 1) |
(→Solution 2) |
||
Line 15: | Line 15: | ||
Then, since the value of n doesn't matter in the quotient <math>\frac{S_{3n}}{S_n}</math>, we can say that | 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> | <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{ | + | 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> |
Solving for <math>a</math>, we get that <math>a=1</math>. | Solving for <math>a</math>, we get that <math>a=1</math>. | ||
Revision as of 07:45, 28 October 2023
Contents
[hide]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,
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
.
Now, we proceed similar to Solution 1 and get that .
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
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.