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Difference between revisions of "2016 AIME II Problems/Problem 9"

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m (→‎Solution 2)
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== Solution 2==
 
== Solution 2==
  
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Using the same reasoning (<math>100</math> isn't very big), we can guess which terms will work. The first case is <math>k=3</math>, so the second and fourth terms of <math>c</math> are <math>100</math> and <math>1000</math>. We let <math>r</math> be the common ratio of the geometric sequence and write the arithmetic relationships in terms of <math>r</math>.
+
Using the same reasoning (<math>100</math> isn't very big), we can guess which terms will work. The first case is <math>k=3</math>, so we assume the second and fourth terms of <math>c</math> are <math>100</math> and <math>1000</math>. We let <math>r</math> be the common ratio of the geometric sequence and write the arithmetic relationships in terms of <math>r</math>.
  
 
The common difference is <math>100-r - 1</math>, and so we can equate: <math>2(99-r)+100-r=1000-r^3</math>. Moving all the terms to one side and the constants to the other yields
 
The common difference is <math>100-r - 1</math>, and so we can equate: <math>2(99-r)+100-r=1000-r^3</math>. Moving all the terms to one side and the constants to the other yields

Revision as of 21:39, 9 April 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

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

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