Solving for a geometric series

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Xiaodan Leng 5 years, 9 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 $\var{s_1}$ 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	2
s_1	integer	25
s_2	integer	50
r	decimal	dec("1")
a	decimal	dec("infinity")

Automatically regenerate variables?

Name

Data type

Value

A random number between and (inclusive) with step size

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

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Parts

Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
  Add

Gap-fill

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

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

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

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

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

$a=\var{a}$

Give a worked solution to the whole question.

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There was an error which means the tests can't run:

Numbas Error: Part a gap 1: [DecimalError] Invalid argument: infinity

Part	Test	Passed?
Gap-fill
Gap-fill		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Number entry
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Taxonomy: mathcentre

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Xiaodan Leng 5 years, 9 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

History

Comment

Checkpoint description

Xiaodan Leng 5 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`