Sequence Property Analysis
A sequence of positive integers with $ a_1 = 1 $ and $ a_9+a_{10}=646 $ is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all $ n\ge1 $, the terms $ a_{2n-1} $, $ a_{2n} $, and $ a_{2n+1} $ are in geometric progression, and the terms $ a_{2n} $, $ a_{2n+1} $, and $ a_{2n+2} $ are in arithmetic progression. Let $ a_n $ be the greatest term in this sequence that is less than 1000. Find $ n+a_n $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Let $ r = a_2 $. Then the first few terms are
\begin{align*}
a_1 &= 1, \\
a_2 &= r, \\
a_3 &= \frac{a_2^2}{a_1} = r^2, \\
a_4 &= 2a_3 - a_2 = 2r^2 - r = r(2r - 1), \\
a_5 &= \frac{a_4^2}{a_3} = \frac{r^2 (2r - 1)^2}{r^2} = (2r - 1)^2, \\
a_6 &= 2a_5 - a_4 = (2r - 1)^2 - r(2r - 1) = (2r - 1)(3r - 2), \\
a_7 &= \frac{a_6^2}{a_5} = \frac{(2r - 1)^2 (3r - 2)^2}{(2r - 1)^2} = (3r - 2)^2, \\
a_8 &= 2a_7 - a_6 = 2(3r - 2)^2 - (2r - 1)(3r - 2) = (3r - 2)(4r - 3), \\
a_9 &= \frac{a_8^2}{a_7} = \frac{(3r - 2)^2 (4r - 3)^2}{(3r - 2)^2} = (4r - 3)^2, \\
a_{10} &= 2a_9 - a_8 = 2(4r - 3)^2 - (3r - 2)(4r - 3) = (4r - 3)(5r - 4).\end{align*}and so on.
More generally, we can prove by induction that
\begin{align*}
a_{2k} &= [(k - 1)r - (k - 2)][kr - (k - 1)], \\
a_{2k + 1} &= [kr - (k - 1)]^2
\end{align*}for any positive integer $ k $.
Then $ (4r - 3)^2 + (4r - 3)(5r - 4) = 646 $. This simplifies to $ 36r^2 - 55r - 625 = 0, $ which factors as $ (r - 5)(36r + 125) = 0 $. Hence, $ r = 5 $.
Then using the formulas above, we can compute that $ a_{16} = 957 $ and $ a_{17} = 1089, $ so the final answer is $ 16 + 957 = \boxed{973} $.
Freeflows