Solving for a geometric series #2

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Frank Doheny

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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

Gave some feedback: Ready to use

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Frank Doheny 8 years, 4 months ago

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Frank Doheny 8 years, 4 months ago

Created this as a copy of Solving for a geometric series.

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 $\var{s1}$ and the sum to infinity of the series is $\var{s2}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

n	integer	5
s1	integer	40
s2	integer	75
r	decimal	dec("8.586207034844889158323794930253590845512e-1")
a	number	10.6034472387

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Generated value: `integer`

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Parts

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1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Ask the student a question, and give any hints about how they should answer this part.

Determine the value of the common ratio. $r$ = Gap 0.

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

Advice

$S_{\var{n}}=\frac{a(1-r^{\var{n}})}{1-r}=\var{s1}$

$S_{\infty}=\frac{a}{1-r}=\var{s2}$

If we divide one by the other we get:

$\frac{S_{\simplify{n}}}{S_{\infty}}=\frac{\frac{a(1-r^{\simplify{n}})}{1-r}}{\frac{a}{1-r}}=\frac{\var{s1}}{\var{s2}}$

$\frac{S_{\simplify{n}}}{S_{\infty}}=\frac{a(1-r^{\simplify{n}})}{1-r}*\frac{1-r}{a}=\frac{\var{s1}}{\var{s2}}$

$1-r^{{n}}=\frac{\var{s1}}{\var{s2}}$

$1-\frac{\var{s1}}{\var{s2}}=r^{{n}}$

In this example $n=\var{n}$

$r^{\var{n}}=\simplify{({s2}-{s1})/{s2}}$

$r=(\simplify{({s2}-{s1})/{s2}})^{1/\var{n}}$

$r=\simplify{(({s2}-{s1})/{s1})^{1/{n}}}=\var{r}$

Recall $S_{\infty}=\frac{a}{1-r}=\var{s2}$

$a=\var{s2}*(1-{r})$

Inserting the value for $r$ in this equation gives

$a=\var{s2}*\simplify{(1-{r})}$

$a=\var{a}$

Give a worked solution to the whole question.

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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 #2

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`