2016 AIME II Problems/Problem 9

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 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 to get $\boxed{262}$ from there.

Solution by rocketscience

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2016 AIME II Problems/Problem 9

Solution

Solution 2