Simon's copy of Solving for a geometric series #3

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Simon Thomas 6 years, 1 month ago

Created this as a copy of Solving for a geometric series #3.

Name	Status	Author	Last Modified
Solving for a geometric series #3	Ready to use	Frank Doheny	20/03/2018 08:26
Simon's copy of Solving for a geometric series #3	draft	Simon Thomas	25/02/2019 11:35
Solving for a geometric series #3	draft	Xiaodan Leng	11/07/2019 00:26

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Question statement

The second term in a geometric series is $\var{t2}$ and the sum to infinity of the series is $\var{s}$ .

There are two possible series that possess these attributes.

Give any introductory information the student needs.

			Name	Type	Generated Value

t2	integer	1
s	integer	40
r_1	number	0.974341649
r_2	number	0.025658351
a_1	number	1.026334039
a_2	number	38.973665961

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1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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3. Gap Gap 2. Number entry
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4. Gap Gap 3. Number entry
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Ask the student a question, and give any hints about how they should answer this part.

Calculate the value of the larger common ratio. $r$ = Gap 0.

Determine the value of the first term of the series corresponding to this common ratio. $a$ = Gap 1.

Calculate the value of the smaller common ratio. $r$ = Gap 2.

Determine the value of the first term of the series corresponding to this common ratio. $a$ = Gap 3.

Advice

The second term of a geometric series is given by the formula $T_2=ar$ and the sum to infinity of a geometric series is $S_\infty=\frac{a}{1-r}$

$T_2=ar=\var{t2}$

$a=\frac{\var{t2}}{r}$

We can substitute this in for $a$ in the second equation

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

$\frac{\frac{\var{t2}}{r}}{1-r}=\var{s}$

$\frac{\var{t2}}{r}=\var{s}(1-{r})$

$\frac{\var{t2}}{r}=\var{s}-\var{s}{r}$

$\var{t2}=\var{s}r-\var{s}r^2$

$\var{s}r^2-\var{s}r+\var{t2}=0$

This is a quadratic equation which we can solve by formula.

$r=\frac{\var{s}\pm \sqrt{(-\var{s})^2-4\times(\var{s})\times(\var{t2})}}{2\times(\var{s})}$

$r=\frac{\var{s}+\sqrt{\simplify{{s}^2-4*{s}*{t2}}}}{\simplify{2*{s}}}$ or $r=\frac{\var{s}-\sqrt{\simplify{{s}^2-4*{s}*{t2}}}}{\simplify{2*{s}}}$

$r=\frac{\var{s}+\simplify{({s}^2-4*{s}*{t2})^0.5}}{\simplify{2*{s}}}$ or $r=\frac{\var{s}-\simplify{({s}^2-4*{s}*{t2})^0.5}}{\simplify{2*{s}}}$

$r=$ {({s}+({s}^2-4*{s}*{t2})^0.5)/(2*{s})} or $r=$ {({s}-({s}^2-4*{s}*{t2})^0.5)/(2*{s})}

$a=\frac{\var{t2}}{r}$

$a=$ {(2*{s}*{t2})/({s}+({s}^2-4*{s}*{t2})^0.5)} or $a=$ {(2*{s}*{t2})/({s}-({s}^2-4*{s}*{t2})^0.5)}

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This question is used in the following exam:

Arithmetic and Geometric Series by Simon Thomas in Maths support.

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Simon's copy of Solving for a geometric series #3

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Simon Thomas 6 years, 1 month ago

Error in variable testing condition

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Generated value: `integer`

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Gap-fill

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Simon's copy of Solving for a geometric series #3

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 6 years, 1 month ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`