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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>.
  
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==Solution==
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==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 <math>b_2</math>. When we get to <math>b_2=9</math> and <math>a_2=91</math>, we have <math>a_4=271</math> and <math>b_4=729</math>, which works, therefore, the answer is <math>b_3+a_3=81+181=\boxed{262}</math>.
 
Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for <math>b_2</math>. When we get to <math>b_2=9</math> and <math>a_2=91</math>, we have <math>a_4=271</math> and <math>b_4=729</math>, which works, therefore, the answer is <math>b_3+a_3=81+181=\boxed{262}</math>.
  
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Solution by rocketscience
 
Solution by rocketscience
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== See also ==
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{{AIME box|year=2016|n=II|num-b=8|num-a=10}}

Revision as of 21:27, 16 May 2016

The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$. There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$. Find $c_k$.

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 $b_2$. When we get to $b_2=9$ and $a_2=91$, we have $a_4=271$ and $b_4=729$, which works, therefore, the answer is $b_3+a_3=81+181=\boxed{262}$.

Solution by Shaddoll

Solution 2

Using the same reasoning ($100$ isn't very big), we can guess which terms will work. The first case is $k=3$, so we assume the second and fourth terms of $c$ are $100$ and $1000$. We let $r$ be the common ratio of the geometric sequence and write the arithmetic relationships in terms of $r$.

The common difference is $100-r - 1$, and so we can equate: $2(99-r)+100-r=1000-r^3$. Moving all the terms to one side and the constants to the other yields

$r^3-3r = 702$, or $r(r^2-3) = 702$. Simply listing out the factors of $702$ shows that the only factor $3$ less than a square that works is $78$. Thus $r=9$ and we solve from there to get $\boxed{262}$.

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
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