Geometric Arithmetic Sequence
A strictly increasing sequence of positive integers $ a_1 $, $ a_2 $, $ a_3 $, $ \dots $ has the property that for every positive integer $ k $, the subsequence $ a_{2k-1} $, $ a_{2k} $, $ a_{2k+1} $ is geometric and the subsequence $ a_{2k} $, $ a_{2k+1} $, $ a_{2k+2} $ is arithmetic. Suppose that $ a_{13} = 2016 $. Find $ a_1 $.
- 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 $ \frac{a_2}{a_1} = \frac{b}{a}, $ where $ a $ and $ b $ are relatively prime positive integers, and $ a < b $. Then $ a_2 = \frac{b}{a} \cdot a_1, $ and
\[a_3 = \frac{a_2^2}{a_1} = \frac{(b/a \cdot a_1)^2}{a_1} = \frac{b^2}{a^2} \cdot a_1.\]This implies $ a_1 $ is divisible by $ a^2 $. Let $ a_1 = ca^2 $; then $ a_2 = cab, $ $ a_3 = cb^2, $
\begin{align*}
a_4 &= 2a_3 - a_2 = 2cb^2 - cab = cb(2b - a), \\
a_5 &= \frac{a_4^2}{a_3} = \frac{[cb(2b - a)]^2}{(cb^2)} = c(2b - 2a)^2, \\
a_6 &= 2a_5 - a_4 = 2c(2b - a)^2 - cb(2b - a) = c(2b - a)(3b - 2a), \\
a_7 &= \frac{a_6^2}{a_5} = \frac{[c(2b - a)(3b - 2a)]^2}{c(2b - a)^2} = c(3b - 2a)^2, \\
a_8 &= 2a_7 - a_6 = 2c(3b - 2a)^2 - c(2b - a)(3b - 2a) = c(3b - 2a)(4b - 3a), \\
a_9 &= \frac{a_8^2}{a_7} = \frac{[c(3b - 2a)(4b - 3a)]^2}{[c(3b - 2a)^2} = c(4b - 3a)^2,
\end{align*}and so on.
More generally, we can prove by induction that
\begin{align*}
a_{2k} &= c[(k - 1)b - (k - 2)a][kb - (k - 1)a], \\
a_{2k + 1} &= c[kb - (k - 1)a]^2,
\end{align*}for all positive integers $ k $.
Hence, from $ a_{13} = 2016, $
\[c(6b - 5a)^2 = 2016 = 2^5 \cdot 3^2 \cdot 7 = 14 \cdot 12^2.\]Thus, $ 6b - 5a $ must be a factor of 12.
Let $ n = 6b - 5a $. Then $ a < a + 6(b - a) = n, $ and
\[n - a = 6b - 6a = 6(b - a),\]so $ n - a $ is a multiple of 6. Hence,
\[6 < a + 6 \le n \le 12,\]and the only solution is $ (a,b,n) = (6,7,12) $. Then $ c = 14, $ and $ a_1 = 14 \cdot 6^2 = \boxed{504} $.
Freeflows