Solving for a geometric series

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Frank Doheny

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Ready to use

7 years ago

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Frank Doheny 7 years ago

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Frank Doheny 8 years, 4 months ago

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Name	Status	Author	Last Modified
Solving for a geometric series	Ready to use	Frank Doheny	20/03/2018 08:27
Solving for a geometric series #2	Ready to use	Frank Doheny	20/03/2018 08:27
Andrew's copy of Solving for a geometric series	draft	Andrew Dunbar	30/06/2017 22:04
Simon's copy of Solving for a geometric series	draft	Simon Thomas	25/02/2019 11:28
Simon's copy of Solving for a geometric series #2	draft	Simon Thomas	25/02/2019 11:31
Solving for a geometric series (simple)	draft	Simon Thomas	25/02/2019 12:33
Solving for a geometric series	draft	Xiaodan Leng	11/07/2019 00:25
Solving for a geometric series #2	draft	Xiaodan Leng	11/07/2019 00:25

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Question statement

The sum of the first $\var{n}$ terms of a geometric series is $\var{s_1}$ and the sum of the first $\simplify{2*{n}}$ terms is $\var{s_2}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

n	integer	5
s_1	integer	12
s_2	integer	43
r	decimal	dec("1.209027250328989849819442991032451950008")
a	decimal	dec("1.584206528809186357894736842105263157895")

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Generated value: `integer`

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1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Determine the value of the common ratio. $r$ = Gap 0.

Calculate the value of the first term. $a$ = Gap 1.

Advice

$S_{\var{n}}=\frac{a(1-r^{\var{n}})}{1-r}=\var{s_1}$

$S_{\simplify{2*{n}}}=\frac{a(1-r^{\simplify{2*{n}}})}{1-r}=\var{s_2}$

If we divide one by the other we get:

$\frac{S_{\simplify{2*{n}}}}{S_{\var{n}}}=\frac{\frac{a(1-r^{\simplify{2*{n}}})}{1-r}}{\frac{a(1-r^{\var{n}})}{1-r}}=\frac{\var{s_2}}{\var{s_1}}$

$\frac{S_{\simplify{2*{n}}}}{S_{\var{n}}}=\frac{a(1-r^{\simplify{2*{n}}})}{1-r}*\frac{1-r}{a(1-r^{\var{n}})}=\frac{\var{s_2}}{\var{s_1}}$

$\frac{1-r^{\simplify{2*{n}}}}{1-r^{\var{n}}}=\frac{\var{s_2}}{\var{s_1}}$

$\frac{(1-r^\var{n})(1+r^{\var{n}})}{1-r^{\var{n}}}=\frac{\var{s_2}}{\var{s_1}}$

$1+r^{\var{n}}=\frac{\var{s_2}}{\var{s_1}}$

$r^{\var{n}}=\frac{\var{s_2}}{\var{s_1}}-1$

$r^{\var{n}}=\simplify{{s_2}/{s_1}-1}$

$r=\simplify{(({s_2})/{s_1}-1)^{1/{n}}}$

$r=\simplify{(({s_2}-{s_1})/{s_1})^{1/{n}}}=\var{r}$

Recall $S_{\var{n}}=\frac{a(1-r^{\var{n}})}{1-r}=\var{s_1}$

$a=\frac{\var{s_1}*(1-{r})}{1-r^{\var{n}}}$

Inserting the value for $r$ in this equation gives

$a=\frac{\var{s_1}*(\simplify{(1-{r})})}{\simplify{{1-r^{{n}}}}}$

$a=\var{a}$

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Taxonomy: mathcentre

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Frank Doheny 7 years ago

Frank Doheny 8 years, 4 months ago

Frank Doheny 8 years, 4 months ago

Error in variable testing condition

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Generated value: `integer`

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Solving for a geometric series

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Frank Doheny 7 years ago

Frank Doheny 8 years, 4 months ago

Frank Doheny 8 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`