1. ## Geometric sequence problem

"Given $\displaystyle a_{n} = 5, r = \frac{1}{2}$, and $\displaystyle S_{n} = 155$ for a geometric sequence, find $\displaystyle a_{1}$ and $\displaystyle n.$"

Using the formulas $\displaystyle a_{n} = a_{1}(r)^{n-1}$ and $\displaystyle S_{n} = \frac{a_{1} (1- r^{n})}{1 - r}$, I start with these two equations:

$\displaystyle 5 = a_{1}\frac{1}{2}^{n-1}$ and

$\displaystyle 155 = \frac{a_{1}(1 - \frac{1}{2}^{n})}{1 - \frac{1}{2}}$

Simplifying each for $\displaystyle a_{1}$, I get

$\displaystyle a_{1} = \frac{5}{\frac{1}{2}^{n-1}}$ and

$\displaystyle a_{1} = \frac{155}{2(1 - \frac{1}{2}^{n})}$

Therefore,

$\displaystyle \frac{5}{\frac{1}{2}^{n-1}} = \frac{155}{2(1 - \frac{1}{2}^{n})}$

I'm stuck at this point... how do I finish simplifying this equation? I'm guessing it'll involve logs... Any help would be greatly appreciated.

2. Originally Posted by earachefl
"Given $\displaystyle a_{n} = 5, r = \frac{1}{2}$, and $\displaystyle S_{n} = 155$ for a geometric sequence, find $\displaystyle a_{1}$ and $\displaystyle n.$"

Using the formulas $\displaystyle a_{n} = a_{1}(r)^{n-1}$ and $\displaystyle S_{n} = \frac{a_{1} (1- r^{n})}{1 - r}$, I start with these two equations:

$\displaystyle 5 = a_{1}\frac{1}{2}^{n-1}$ and

$\displaystyle 155 = \frac{a_{1}(1 - \frac{1}{2}^{n})}{1 - \frac{1}{2}}$

Simplifying each for $\displaystyle a_{1}$, I get

$\displaystyle a_{1} = \frac{5}{\frac{1}{2}^{n-1}}$ and

$\displaystyle a_{1} = \frac{155}{2(1 - \frac{1}{2}^{n})}$

Therefore,

$\displaystyle \frac{5}{\frac{1}{2}^{n-1}} = \frac{155}{2(1 - \frac{1}{2}^{n})}$

I'm stuck at this point... how do I finish simplifying this equation? I'm guessing it'll involve logs... Any help would be greatly appreciated.
no logarithms needed here, just some "simple" simplification

you seem to be on the money so far. i will restart the problem though. you way is fine, but it seems having the halves around was confusing you. i will rewrite the problem using powers of 2, not 1/2

$\displaystyle a_n = 5, r = \frac 12, \mbox{ and } S_n = 155$

$\displaystyle a_n = ar^{n - 1}$

$\displaystyle \Rightarrow 5 = a \left( \frac 12 \right)^{n - 1} = a \cdot 2^{1 - n}$

$\displaystyle \Rightarrow a = 5 \cdot 2^{n - 1}$ ....................(1)

$\displaystyle S_n = \frac {a \left( 1 - r^n \right)}{1 - r}$

$\displaystyle \Rightarrow 155 = \frac {a \left( 1 - 2^{-n} \right)}{\frac 12}$

$\displaystyle \Rightarrow \frac {155}2 = a \left( 1 - 2^{-n} \right)$

$\displaystyle \Rightarrow a = \frac {155}{2 - 2^{1 - n}}$ .....................(2)

equating (1) and (2), we obtain:

$\displaystyle 5 \cdot 2^{n - 1} = \frac {155}{2 - 2^{1 - n}}$

$\displaystyle \Rightarrow 2^{n - 1} = \frac {31}{2 - 2^{1 - n}}$

$\displaystyle \Rightarrow 2^n - 1 = 31$

$\displaystyle \Rightarrow 2^n = 32$

$\displaystyle \Rightarrow \boxed{n = 5}$

back-substitute into (1):

$\displaystyle \Rightarrow a = 5 \cdot 2^4$

$\displaystyle \Rightarrow \boxed{a = 80}$

if you insist on picking up where you left off:

... $\displaystyle \frac{5}{\frac{1}{2}^{n-1}} = \frac{155}{2(1 - \frac{1}{2}^{n})}$
$\displaystyle \Rightarrow \frac 5{2^{1 - n}} = \frac {155}{2 \left( 1 - 2^{-n} \right)}$ .........divide both sides by 5

$\displaystyle \Rightarrow \frac 1{2^{1 - n}} = \frac {31}{2 \left( 1 - 2^{-n} \right)}$ ............multiply both sides by $\displaystyle 2 \left( 1 - 2^{-n} \right)$

$\displaystyle \Rightarrow \frac {2 \left( 1 - 2^{-n} \right)}{2^{1 - n}} = 31$

$\displaystyle \Rightarrow 2^{1 - (1 - n)} \left( 1 - 2^{-n} \right) = 31$

$\displaystyle \Rightarrow 2^n \left( 1 - 2^{-n} \right) = 31$

$\displaystyle \Rightarrow 2^n - 1 = 31$

and the rest is trivial

3. i will rewrite the problem using powers of 2, not 1/2

$\displaystyle \Rightarrow 5 = a \left( \frac 12 \right)^{n - 1} = a \cdot 2^{1 - n}$

So when you are doing this, effectively, you are doing

$\displaystyle a \frac 12 ^{(-1)(n - 1)(-1)}$? Is that the trick?

4. Originally Posted by earachefl
So when you are doing this, effectively, you are doing

$\displaystyle a \frac 12 ^{(-1)(n - 1)(-1)}$? Is that the trick?
no.
$\displaystyle \frac 12 = 2^{-1}$

so the trick is: $\displaystyle a \left( \frac 12 \right)^{n - 1} = a \left( 2^{-1} \right)^{n - 1} = a \cdot 2^{(-1)(n -1)} = a \cdot 2^{1 - n}$

5. Hello, earachefl!

Another approach . . .

Given: .$\displaystyle a_n \,=\, 5,\;\;r \,=\, \frac{1}{2},\;\;S_n \,=\, 155$ . for a geometric sequence.
Find: $\displaystyle a_1$ and $\displaystyle n$

We have: .$\displaystyle a_n \;=\;a_1\left(\frac{1}{2}\right)^{n-1}\quad\Rightarrow\quad \frac{a_1}{2^{n-1}}\;=\;5$ .[1]

and: .$\displaystyle S_n \;=\;a_1\,\frac{1 - \left(\frac{1}{2}\right)^n}{1-\frac{1}{2}} \;=\;a_1\,\frac{1-\frac{1}{2^n}}{\frac{1}{2}} \;=\;2a_1\left(1-\frac{1}{2^n}\right)\quad\Rightarrow\quad 2a_1 \,- \,\frac{a_1}{2^{n-1}} \;=\;155$ .[2]

Add [1] and [2]: .$\displaystyle 2a_1 \:=\:160\quad\Rightarrow\quad\boxed{ a_1 \:=\:80}$

Substitute into [1]: .$\displaystyle \frac{80}{2^{n-1}} \:=\:5\quad\Rightarrow\quad 2^{n-1} \:=\:16\quad\Rightarrow\quad 2^{n-1} \:=\:2^4$

Therefore: .$\displaystyle n-1\:=\:4\quad\Rightarrow\quad\boxed{n \:=\:5}$