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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: Yendall on September 26, 2012, 06:36:06 pm

Title: Determining 'r' when given Infinite Sum and Sum of Terms
Post by: Yendall on September 26, 2012, 06:36:06 pm: This is really similar to the thread I started ealier, however a lot harder to solve.

I was just doing a NEAP practice exam and came across a question very similar to that of the previous thread:

The sum of the first three terms of a geometric series is 57 while the infinite sum is 81.
The common ratio is:

I know that the infinite sum will be: $S_\infty = \frac{a}{(1 - r)}$ OR $81 = \frac{a}{(1 - r)}$

I know that the sum of n terms will be: $S_3 = \frac{a(1-r^3)}{1-r}$ OR $57 =\frac{a(1-r^3)}{1-r}$

When entered into CAS like this:
$solve(57=\frac{a(1 - r^3)}{1-r},a)$

It returns the answer:
$\frac{-a(1 - r^3)}{r -1} = 57$

Is it okay to conclude that $a = 57$ when presented with an answer like this?

From there I could solve for r by:
$81 = \frac{57}{(1 - r)}$
$solve(81 = \frac{57}{(1-r)},r)$

$r = 0.296296$

However, that's the incorrect solution for 'r'. How do I solve something like this?

Thanks for any help.
Title: Re: Determining 'r' when given Infinite Sum and Sum of Terms
Post by: Phy124 on September 26, 2012, 06:52:25 pm: I didn't do further but I think I've concluded you are trying to find $a$ and $r$ , using the equations

$81=\frac{57}{1-r}$

$57 = \frac{a(1-r^3)}{1-r}$

In which case you can solve them simultaneously on your calculator:

$solve\left(81=\frac{a}{1-r} \ and \ 57 = \frac{a(1-r^3)}{1-r},a,r\right)$

$a = 27$

$r = \frac{2}{3}$

Quote from: Yendall on September 26, 2012, 06:36:06 pm
When entered into CAS like this:
$solve(57=\frac{a(1 - r^3)}{1-r},a)$

It returns the answer:
$\frac{-a(1 - r^3)}{r -1} = 57$

Is it okay to conclude that $a = 57$ when presented with an answer like this?

That answer isn't telling you anything, it is just a different form of what you inputted, as you are trying to solve one equation for two variables.

edit: had solved using 91 and not 81, fixed now
Title: Re: Determining 'r' when given Infinite Sum and Sum of Terms
Post by: Yendall on September 26, 2012, 07:09:38 pm: Quote from: rangaaaaaa on September 26, 2012, 06:52:25 pm
I didn't do further but I think I've concluded you are trying to find $a$ and $r$ , using the equations

$81=\frac{57}{1-r}$

$57 = \frac{a(1-r^3)}{1-r}$

In which case you can solve them simultaneously on your calculator:

$solve\left(81=\frac{a}{1-r} \ and \ 57 = \frac{a(1-r^3)}{1-r},a,r\right)$

$a = 27$

$r = \frac{2}{3}$

Quote from: Yendall on September 26, 2012, 06:36:06 pm
When entered into CAS like this:
$solve(57=\frac{a(1 - r^3)}{1-r},a)$

It returns the answer:
$\frac{-a(1 - r^3)}{r -1} = 57$

Is it okay to conclude that $a = 57$ when presented with an answer like this?

That answer isn't telling you anything, it is just a different form of what you inputted, as you are trying to solve one equation for two variables.

edit: had solved using 91 and not 81, fixed now

Thaaaank you!