Geometric series 2

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joshua boddy

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joshua boddy 8 years, 11 months ago

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Created this as a copy of Geometric series 2.

Name	Status	Author	Last Modified
Geometric series 1	Ready to use	joshua boddy	01/06/2016 09:47
Geometric series 2	Ready to use	joshua boddy	01/06/2016 09:47
CF Maths In class test three mock paper question 3	Ready to use	Jinhua Mathias	28/08/2024 12:15
Geometric series 2	Ready to use	Jinhua Mathias	28/08/2024 12:15
Evan's copy of Geometric series 2	draft	Evan Weinberg	22/08/2016 14:43
Evan's copy of Evan's copy of Geometric series 2	draft	Evan Weinberg	24/08/2016 02:38
Geometric series 2	draft	Xiaodan Leng	11/07/2019 06:50
Geometric series 1	draft	Xiaodan Leng	11/07/2019 06:53

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Question statement

For the geometric series:

$\var{t1} - \simplify{{t1}/{td}} + \simplify{{t1}/{td}^2} - \simplify{{t1}/{td}^3} +...$

Give any introductory information the student needs.

			Name	Type	Generated Value

t1	integer	176
tr	integer	6
tsum	integer	18
td	integer	4

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Name

Data type

Value

xxxxxxxxxx
 
td*random(10..100)

JME function reference

Description

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

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Generated value: `integer`

176

← Depends on:

This variable doesn't seem to be used anywhere.

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Part a)

To find the $i^{th}$ term for a specific $i$ (in this case $\var{tr}$ ) we need to recall the following formulae:

$\displaystyle\sum\limits_{i=1}^nar^{i-1}$ $_{..(I)}$

$a_i=ar^{i-1}$ $_{..(II)}$

$a$ is the first term in the sequence. Namely, $\var{t1}$ .

To find the common ratio, $r$ , simply divide one term by its predecessor.

The known values for $a$ , $r$ and $i$ are then subsituted into $_{(II)}$ .

Part b)

To find the general $i^{th}$ term, $a_i$ , formula of a geometric series, we substitute known values of $a$ and $r$ into $_{(II)}$ to produce the formula in $i$ .

In this case,

$a_i=\var{t1}(\simplify{-1/{td}})^{i-1}$

Part c)

To find the sum of the first $n$ (in this case $\var{tsum}$ ) terms of a geometric series, we turn to the following formula:

$\displaystyle\sum\limits_{i=1}^nar^{i-1}=a\left(\frac{1-r^n}{1-r}\right)$ $_{..(III)}$

The sum is found by subsituting the known values for $a$ , $r$ and $n$ into $_{(III)}$ .

Part d)

The sum to infinity of a geometric series can be found by first taking the usual sum from $_{(III)}$ and setting $n$ to $\infty$ :

$\displaystyle\sum\limits_{i=1}^{\infty}ar^{i-1}=a\left(\frac{1-r^\infty}{1-r}\right)$ $_{..(III)}$

When $\left|r\right|<0$ , note that $r^\infty$ tends to $0$ . The above formula simplifies:

$a\left(\frac{1-0}{1-r}\right)=a\left(\frac{1}{1-r}\right)=\left(\frac{a}{1-r}\right)$

$\therefore\;\;\;\displaystyle\sum\limits_{i=1}^{\infty}ar^{i-1}=\left(\frac{a}{1-r}\right)$ $_{..(IV)}$

The final sum is calculated by substituting the values for $a$ and $r$ into $_{(IV)}$ .

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This question is used in the following exam:

CF Maths weekly Portfolio 7 - Arithmetic and Geometric sequence by Jinhua Mathias in Core Foundation Maths.

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joshua boddy 8 years, 11 months ago

Deactivated user 9 years, 5 months ago

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Geometric series 2

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Taxonomy: mathcentre

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joshua boddy 8 years, 11 months ago

Deactivated user 9 years, 5 months ago

Error in variable testing condition

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Generated value: integer

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Number entry

Advanced settings

Generated value: `integer`