Sequences Addition Problem
Let $ a_1, $ $ a_2, $ $ a_3, $ $ \dots $ be an arithmetic sequence, and let $ b_1, $ $ b_2, $ $ b_3, $ $ \dots $ be a geometric sequence. The sequence $ c_1, $ $ c_2, $ $ c_3, $ $ \dots $ has $ c_n = a_n + b_n $ for each positive integer $ n $. If $ c_1 = 1, $ $ c_2 = 4, $ $ c_3 = 15, $ and $ c_4 = 2, $ compute $ c_5 $.
- 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 the arithmetic sequence be $ a_n = a + (n - 1)d, $ and let the geometric sequence be $ b_n = br^{n-1} $. Then
\begin{align*}
a + b &= 1, \\
a + d + br &= 4, \\
a + 2d + br^2 &= 15, \\
a + 3d + br^3 &= 2.\end{align*}Subtracting pairs of equations, we get
\begin{align*}
d + br - b &= 3, \\
d + br^2 - br &= 11, \\
d + br^3 - br^2 &= -13.\end{align*}Again subtracting pairs of equations, we get
\begin{align*}
br^2 - 2br + b &= 8, \\
br^3 - 2br^2 + br &= -24.\end{align*}We can write these as
\begin{align*}
b(r - 1)^2 &= 8, \\
br(r - 1)^2 &= -24.\end{align*}Dividing these equations, we get $ r = -3 $. Then $ 16b = 8, $ so $ b = \frac{1}{2} $. Then
\begin{align*}
a + \frac{1}{2} &= 1, \\
a + d - \frac{3}{2} &= 4.\end{align*}Solving for $ a $ and $ d, $ we find $ a = \frac{1}{2} $ and $ d = 5 $.
Hence,
\begin{align*}
c_5 &= a_5 + b_5 \\
&= a + 4d + br^4 \\
&= \frac{1}{2} + 4 \cdot 5 + \frac{1}{2} \cdot (-3)^4 \\
&= \boxed{61}.\end{align*}
Freeflows