Difference between revisions of "2022 AMC 10B Problems/Problem 15"
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Solving for <math>a</math>, we get that <math>a=1</math>. | Solving for <math>a</math>, we get that <math>a=1</math>. | ||
− | + | Since the sum of the first <math>n</math> odd number is <math>n^2</math>, <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>. | |
==Solution 3 (Quick Insight)== | ==Solution 3 (Quick Insight)== |
Revision as of 07:49, 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 .
Since the sum of the first odd number 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
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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