Difference between revisions of "2022 AMC 10B Problems/Problem 15"
Kavya.rajesh (talk | contribs) |
(→Solution 3) |
||
Line 30: | Line 30: | ||
We can simplify further to get <math>(a+2)/a</math>=<math>(a+5)/(a+1)</math> | We can simplify further to get <math>(a+2)/a</math>=<math>(a+5)/(a+1)</math> | ||
Solving for <math>a</math>, we get that <math>a=1</math>. Now, we proceed similar to the previous solutions and get that <math>S_n = \fbox{D. 400}</math> | Solving for <math>a</math>, we get that <math>a=1</math>. Now, we proceed similar to the previous solutions and get that <math>S_n = \fbox{D. 400}</math> | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | https://youtu.be/7ztNpblm2TY | ||
+ | |||
+ | ~Education, the Study of Everything | ||
== 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 19:45, 19 November 2022
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, the value of is
~mathboy100
Solution 2 (Quick Insight)
Recall that the sum of the first odd numbers is .
. Thus
~numerophile
Solution 3
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 =
We can simplify further to get = Solving for , we get that . Now, we proceed similar to the previous solutions and get that
Video Solution 1
~Education, the Study of Everything
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.