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 16:18, 22 March 2018
Contents
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 | ||
