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maths help please weird question

PART ii please thanks

maths help please.PNG

maths help please.PNG

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Original post by h26

PART ii please thanks

You got two

a

values and two

r

values.

Say these are

a_1, a_2

and

r_1 , r_2

. (because I can't be bothered to do part a and find out what they are)

Then we have two geo progressions:

u_n = a_1r_1^{n-1}

and

v_n = a_2 r_2^{n-1}

You want to show that

\dfrac{u_n}{v_n}

gives the ratio

2^{n-2} : 3^{n-2}

.

OP

Original post by RDKGames

You got two

a

values and two

r

values.

Say these are

a_1, a_2

and

r_1 , r_2

. (because I can't be bothered to do part a and find out what they are)

Then we have two geo progressions:

u_n = a_1r_1^{n-1}

and

v_n = a_2 r_2^{n-1}

You want to show that

\dfrac{u_n}{v_n}

gives the ratio

2^{n-2} : 3^{n-2}

.

Thanks but I can't seem to simplify it down - not sure what to do

I've got the first marking point and that's it

InnocentJackski

2+2 = 4
4-1 = 3
Quick maths

OP

Original post by RDKGames

You got two

a

values and two

r

values.

Say these are

a_1, a_2

and

r_1 , r_2

. (because I can't be bothered to do part a and find out what they are)

Then we have two geo progressions:

u_n = a_1r_1^{n-1}

and

v_n = a_2 r_2^{n-1}

You want to show that

\dfrac{u_n}{v_n}

gives the ratio

2^{n-2} : 3^{n-2}

.

Thanks but I can't seem to simplify it down - not sure what to do

I've got the first marking point and that's it

OP

Bump

Study Forum Helper

Original post by h26

Thanks but I can't seem to simplify it down - not sure what to do

I've got the first marking point and that's it

So then you have

\dfrac{u_n}{v_n} = \dfrac{10 \times \left( \frac{3}{5} \right)^{n-1}}{15 \times \left( \frac{2}{5} \right)^{n-1}}

.

Simplify this down to either

\dfrac{2^{n-2}}{3^{n-2}}

or

\dfrac{3^{n-2}}{2^{n-2}}

- either one implies the wanted ratio.

OP

Original post by RDKGames

So then you have

\dfrac{u_n}{v_n} = \dfrac{10 \times \left( \frac{3}{5} \right)^{n-1}}{15 \times \left( \frac{2}{5} \right)^{n-1}}

.

Simplify this down to either

\dfrac{2^{n-2}}{3^{n-2}}

or

\dfrac{3^{n-2}}{2^{n-2}}

- either one implies the wanted ratio.

So got the top part but not sure how you simplified it down

Study Forum Helper

Original post by h26

So got the top part but not sure how you simplified it down

Requires basic index laws and fraction manipulation really.

Try to have a go at it and post what you end up with if you cant get it.

OP

Original post by RDKGames

Requires basic index laws and fraction manipulation really.

Try to have a go at it and post what you end up with if you cant get it.

Had a go-not working

Original post by InnocentJackski

2+2 = 4
4-1 = 3
Quick maths

You're not funny

OP

Original post by RDKGames

Requires basic index laws and fraction manipulation really.

Try to have a go at it and post what you end up with if you cant get it.

Here is a better pic

Original post by h26

Here is a better pic

The bits where you have (2/5)^n multiplied by (2/5)^-1 can be simplified to (2/5)^(n-1), same applies to the (3/5)^n and (3/5)^-1 stuff on the first few lines you wrote

IF THAT IS ANY HELP idk if it will be

Study Forum Helper

Original post by h26

Here is a better pic

It's valid but completely missed the point.

We want to get all the 2s together, and all the 3s together. So splitting the fraction on your first line is going against this point.

\dfrac{10 \times \left( \frac{3}{5} \right)^{n-1}}{15 \times \left( \frac{2}{5} \right)^{n-1}} = \dfrac{2 \times 3^{n-1}}{3 \times 2^{n-1}}

as a massive starter.

OP

Original post by RDKGames

It's valid but completely missed the point.

We want to get all the 2s together, and all the 3s together. So splitting the fraction on your first line is going against this point.

\dfrac{10 \times \left( \frac{3}{5} \right)^{n-1}}{15 \times \left( \frac{2}{5} \right)^{n-1}} = \dfrac{2 \times 3^{n-1}}{3 \times 2^{n-1}}

as a massive starter.

I am really confused :frown:

:frown:

Study Forum Helper

Original post by h26

I am really confused :frown:

:frown:

\dfrac{10 \times \left( \frac{3}{5} \right)^{n-1}}{15 \times \left( \frac{2}{5} \right)^{n-1}} = \dfrac{2\times 5 \times \frac{3^{n-1}}{5^{n-1}}}{3 \times 5 \times \frac{2^{n-1}}{5^{n-1}}} = \dfrac{2 \times 3^{n-1}}{3 \times 2^{n-1}}

The 5's cancel on numerator and denominator, then mult top and bottom by

5^{n-1}

Study Forum Helper

Original post by FragHMM

is this edexcel gcse?

No.

Is this GCSEs by any chance, because if it is, I'm completely screwed... I barely understand the question, nevermind the working out!

Original post by FragHMM

is this edexcel gcse?

Original post by RDKGames

No.

Original post by Rachana.L

Is this GCSEs by any chance, because if it is, I'm completely screwed... I barely understand the question, nevermind the working out!

:rofl:

Study Forum Helper

Original post by Rachana.L

Is this GCSEs by any chance, because if it is, I'm completely screwed... I barely understand the question, nevermind the working out!

I see these questions on A-Level threads often enough to make me question why GCSE students don't know their own spec that they're doing...

Even if you haven't looked at the spec, you should know that it's not in GCSE because your teacher hasn't taught it, or because it hasn't come up in any practice questions you've (hopefully) been doing!

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