Difference between revisions of "2013 AMC 12A Problems/Problem 14"
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If we factor the two known terms we get <math>162=2\cdot 3^4</math> and <math>1250=2\cdot 5^4</math>, thus the quotient is obviously <math>5/3</math> and therefore <math>x=162\cdot(5/3) = 270</math>. | If we factor the two known terms we get <math>162=2\cdot 3^4</math> and <math>1250=2\cdot 5^4</math>, thus the quotient is obviously <math>5/3</math> and therefore <math>x=162\cdot(5/3) = 270</math>. | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/RdIIEhsbZKw?t=944 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == |
Latest revision as of 03:32, 21 January 2021
Problem
The sequence
, , , ,
is an arithmetic progression. What is ?
Solution 1
Since the sequence is arithmetic,
+ = , where is the common difference.
Therefore,
= - = , and
= () =
Now that we found , we just add it to the first term to find :
+ =
= = = = , which is
Solution 2
As the sequence , , , , is an arithmetic progression, the sequence must be a geometric progression.
If we factor the two known terms we get and , thus the quotient is obviously and therefore .
Video Solution
https://youtu.be/RdIIEhsbZKw?t=944
~ pi_is_3.14
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 12 Problems and Solutions |
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