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Aesity's question at Yahoo! Answers regarding a geometric sequence

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MarkFL

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Feb 24, 2012
13,775
Here is the question:

Geometric sequence question?

The first term of a geometric sequence is 27 and the sixth term is 512/9.

1) Find the seventh term

I have problem with this working. Either the calculator is pulling a false one on me or IDK.
Since 512/9 = 27r^(6-1)
then (512/9) / 27 = r^5
r^5=512/243
r= (512/243) ^ 1/5
What I got for r was 0.421399177

To arrive at the seventh term; 27(0.421399177)^(7-1). I got 0.15119
This is no where close to the answer but apparently there doesn't appear to be any anomalies with my workings.





2) Find the sum of the first five terms of this sequence.

I need explanations behind the workings. Thanks!
Here is a link to the question:

Geometric sequence question? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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MarkFL

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Feb 24, 2012
13,775
Re: Aesity's question at Yahoo! Ansers regarding a geometric sequence

Hello Aesity,

We are given:

$\displaystyle a_1=27$

$\displaystyle a_6=\frac{512}{9}$

The $n$th term in the sequence is:

$\displaystyle a_n=27r^{n-1}$

and so we may determine $r$ by writing:

$\displaystyle a_6=\frac{512}{9}=27r^{6-1}$

$\displaystyle r^5=\frac{2^9}{3^5}$

$\displaystyle r=\frac{2^{\frac{9}{5}}}{3}=\frac{2\cdot2^{\frac{4}{5}}}{3}$

and so we have:

$\displaystyle a_n=27\left(\frac{2\cdot2^{\frac{4}{5}}}{3} \right)^{n-1}$

1.) Thus, we may now answer the first question:

$\displaystyle a_7=27\left(\frac{2\cdot2^{\frac{4}{5}}}{3} \right)^{6}=\frac{64\cdot2^{\frac{24}{5}}}{27}= \frac{1024 \cdot 2^{\frac{4}{5}}}{27}$

To find the sum of the first $n$ terms, consider:

$\displaystyle S_n=a_1+a_2+a_3+\cdots+a_n$

$\displaystyle S_n=a_1+a_1r+a_1r^2+\cdots+a_1r^{n-1}$

Multiply through by $r$:

$\displaystyle rS_n=a_1r+a_1r^2+a_1r^3+\cdots+a_1r^{n}=S_n+a_1r^{n}-a_1$

$\displaystyle rS_n-S_n=a_1r^{n}-a_1$

$\displaystyle (r-1)S_n=a_1r^{n}-a_1$

$\displaystyle S_n=\frac{a_1r^{n}-a_1}{r-1}=a_1\cdot\frac{r^n-1}{r-1}$

2.) Now we may answer the second question:

$\displaystyle S_5=a_1\cdot\frac{r^5-1}{r-1}=27\cdot\frac{\left(\frac{2\cdot2^{\frac{4}{5}}}{3} \right)^5-1}{\frac{2\cdot2^{\frac{4}{5}}}{3}-1}=\frac{269}{3(2\cdot2^{\frac{4}{5}}-3)}$
 
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