Jinhua's copy of Arithmetic Series 4

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Created this as a copy of Arithmetic Series 4.

Name	Status	Author	Last Modified
Arithmetic Series 4	Ready to use	joshua boddy	01/06/2016 09:47
Jinhua's copy of Arithmetic Series 4	Ready to use	Jinhua Mathias	28/08/2024 12:15
Evan's copy of Arithmetic Series 4	draft	Evan Weinberg	24/08/2016 02:34
Sequences 3	draft	Mark Hodds	06/11/2016 20:42
Arithmetic Series 4	draft	Xiaodan Leng	11/07/2019 05:32
Arithmetic Series 4	draft	Xiaodan Leng	11/07/2019 07:05

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Question statement

The $\var{n}$ ^thterm of an arithmetic series is $\simplify{{t1}+({n}-1)*{d}}$

The common difference is $\var{d}$

Give any introductory information the student needs.

			Name	Type	Generated Value

t1	integer	5
d	integer	4
t2	integer	9
t3	integer	13
t4	integer	17
tr	integer	7
tsum	integer	17
tans	rational	629
n	integer	18
a	rational	2
b	rational	3
bsq	rational	9
dscrm	decimal	dec("8.949301648732151211954765177160623992676e+1")
solvet1	rational	-3/4
solvet2	decimal	dec("2.237325412183037802988691294290155998169e+1")
nc	decimal	dec("21.62")

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b
d
t2
t3
t4
tans

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"Part a)" - Minimum accepted value
"Part a)" - Maximum accepted value
"Part c)" - Minimum accepted value
"Part c)" - Maximum accepted value

Advice

Part a)

To find the firstterm $a_1$ , we need to first start with the formula for the $n^{th}$ term:

$a_n=a_1+d(n-1)$ $_{..(i)}$ which is rearranged in terms of $a_1$ : $a_1=a_n-d(n-1)$ $_{..(ii)}$

We then identify the variables which we know, namely: $n=\var{n}$ , $a_n=\simplify{{t1}+({n}-1)*{d}}$ , $d=\var{d}$

These values can then be subsituted into $_{(ii)}$ and solved for $a_1$ :

$a_1=\simplify{{t1}+({n}-1)*{d}}-\var{d}(\var{n}-1)$

Part b)

To find the sum of the first $\var{tsum}$ terms we take the formula for the sum of an arithmetic series:

$S_n=\frac{n}{2}(a_1+a_n)$ $_{..(iii)}$

While we don't yet know the value of $a_n$ , it is obtainable with formula $_{..(i)}$ . It is common to substitute this formula into the formula for the sum as follows:

$S_n=\frac{n}{2}(a_1+a_1+d(n-1))$

$\therefore\;\;\;S_n=\frac{n}{2}(2a_1+d(n-1))$ $_{..(iv)}$

We identify the variables which we know, namely: $n=\var{tsum}$ , $a_1=\var{t1}$ , $d=\var{d}$

They are substituted into $_{..(iv)}$ and solved for $S_n$ :

$S_\var{tsum}=\frac{\var{tsum}}{2}(2\times\var{t1}+\var{d}(\var{tsum}-1))$

Part c)

To determine the least value of $n$ for which the sum of the first $n$ terms of the series exceeds $1000$ , the sum formula $_{..(iv)}$ is set up with the lower limit value of $1000$ substituted in:

$1000=\frac{n}{2}(2\times\var{t1}+\var{d}(n-1))$

It is then expanded algebraically to give the following quadratic:

$\simplify{{d}/2}n^2+\simplify{{2*{t1}-{d}}/2}n-1000=0$

$n$ can then be found using the quadratic formula:

$n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=\simplify{{d}/2}$ , $b=\simplify{{2*{t1}-{d}}/2}$ and $c=-1000$

Thus,

$n=\frac{-\var{b}\pm\sqrt{(\var{b})^2-\simplify{{-1000*4*{a}}}}}{\simplify{2*{a}}}$

$\therefore\;\;\;n=-\left(\simplify{{2*{t1}-{d}}/{2*2*{a}}}\right)\pm\left(\frac{\sqrt{\simplify{{bsq}-{-1000*4*{a}}}}}{\simplify{2*{a}}}\right)$

In this context, we are talking of positive $n^{th}$ terms, so we disregard the negative square root.

The result is then computed to be: $\var{nc}$

If this is the value of $n$ which produces $1000$ , it follows that the next whole number will be the least value of $n$ for which the sum of the first $n$ terms of the series exceeds $1000$ .

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Jinhua's copy of Arithmetic Series 4

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

joshua boddy 8 years, 9 months ago

Deactivated user 9 years, 3 months ago

Error in variable testing condition

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Generated value: integer

Parts

Number entry

Advanced settings

Generated value: `integer`