Difference between revisions of "2016 AIME II Problems/Problem 9"
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The sequences of positive integers <math>1,a_2, a_3,...</math> and <math>1,b_2, b_3,...</math> are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let <math>c_n=a_n+b_n</math>. There is an integer <math>k</math> such that <math>c_{k-1}=100</math> and <math>c_{k+1}=1000</math>. Find <math>c_k</math>. | The sequences of positive integers <math>1,a_2, a_3,...</math> and <math>1,b_2, b_3,...</math> are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let <math>c_n=a_n+b_n</math>. There is an integer <math>k</math> such that <math>c_{k-1}=100</math> and <math>c_{k+1}=1000</math>. Find <math>c_k</math>. | ||
Revision as of 17:18, 22 March 2018
Contents
[hide]Problem
The sequences of positive integers and
are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let
. There is an integer
such that
and
. Find
.
Solution 1
Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for . When we get to
and
, we have
and
, which works, therefore, the answer is
.
Solution by Shaddoll
Solution 2
Using the same reasoning ( isn't very big), we can guess which terms will work. The first case is
, so we assume the second and fourth terms of
are
and
. We let
be the common ratio of the geometric sequence and write the arithmetic relationships in terms of
.
The common difference is , and so we can equate:
. Moving all the terms to one side and the constants to the other yields
, or
. Simply listing out the factors of
shows that the only factor
less than a square that works is
. Thus
and we solve from there to get
.
Solution by rocketscience
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |