1. ## Geometric sequence problem

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

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

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

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

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

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

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

Therefore,

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

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

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

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

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

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

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

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

Therefore,

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

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

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

$a_n = ar^{n - 1}$

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

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

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

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

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

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

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

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

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

$\Rightarrow 2^n - 1 = 31$

$\Rightarrow 2^n = 32$

$\Rightarrow \boxed{n = 5}$

back-substitute into (1):

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

$\Rightarrow \boxed{a = 80}$

if you insist on picking up where you left off:

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

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

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

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

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

$\Rightarrow 2^n - 1 = 31$

and the rest is trivial

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

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

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

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

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

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

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

5. Hello, earachefl!

Another approach . . .

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

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

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

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

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

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