# Geometric sequences help

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dwrfwrw

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#1

https://gyazo.com/535bd4824e4543f5a7e689a8caf5f5a2

I got the nth term of this sequence but I'm not sure why the number of terms for the last question is 7, I took logs and got it but I'm still not convinced why considering the fact you can't take logs of negative numbers leading me to ignore the negative number?

Please help

I got the nth term of this sequence but I'm not sure why the number of terms for the last question is 7, I took logs and got it but I'm still not convinced why considering the fact you can't take logs of negative numbers leading me to ignore the negative number?

Please help

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mqb2766

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#2

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#2

(Original post by

https://gyazo.com/535bd4824e4543f5a7e689a8caf5f5a2

I got the nth term of this sequence but I'm not sure why the number of terms for the last question is 7, I took logs and got it but I'm still not convinced why considering the fact you can't take logs of negative numbers leading me to ignore the negative number?

Please help

**dwrfwrw**)https://gyazo.com/535bd4824e4543f5a7e689a8caf5f5a2

I got the nth term of this sequence but I'm not sure why the number of terms for the last question is 7, I took logs and got it but I'm still not convinced why considering the fact you can't take logs of negative numbers leading me to ignore the negative number?

Please help

2 = 2^?

...

64 = 2^?

Last edited by mqb2766; 3 months ago

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dwrfwrw

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#3

(Original post by

64 = 2^?

**mqb2766**)64 = 2^?

so therefore 64=-2 to the n-1

but you can't take logs of negative numbers...

i got un= -2 to the n-1 which is right

Last edited by dwrfwrw; 3 months ago

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mqb2766

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#4

(Original post by

Maybe I did something wrong but I got un=ar to the n-1

so therefore 64=-2 to the n-1

but you can't take logs of negative numbers...

**dwrfwrw**)Maybe I did something wrong but I got un=ar to the n-1

so therefore 64=-2 to the n-1

but you can't take logs of negative numbers...

So what power of 2 equals 64.

Last edited by mqb2766; 3 months ago

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dwrfwrw

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#5

(Original post by

You should know a doubling 2 sequence without taking logs?

So what power of 2 equals 64.

**mqb2766**)You should know a doubling 2 sequence without taking logs?

So what power of 2 equals 64.

I know the sequence is doubling but why did you ignore the minus sign?

idk if this works but I tried setting equation of Un to the last term of 64..

Last edited by dwrfwrw; 3 months ago

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mqb2766

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#6

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#6

(Original post by

Yeah it's 6?

**dwrfwrw**)Yeah it's 6?

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#7

(Original post by

Of course and that is the n-1 term so n=7.

**mqb2766**)Of course and that is the n-1 term so n=7.

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#8

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#8

(Original post by

sorry to annoy you again but I know the sequnce is doubling but why did you ignore the minus again? just wanna clear this up

**dwrfwrw**)sorry to annoy you again but I know the sequnce is doubling but why did you ignore the minus again? just wanna clear this up

What role does the - sign have?

Last edited by mqb2766; 3 months ago

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#9

(Original post by

The negative sign occurs when n=2,4,6,... You know r is -2 and the sequence is

(-2)^(n-1)

The next term would be -128 which would correspond to (-2)^7

**mqb2766**)The negative sign occurs when n=2,4,6,... You know r is -2 and the sequence is

(-2)^(n-1)

The next term would be -128 which would correspond to (-2)^7

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#10

am not sure what to do in this question.

I did integral between (f(x)-g(x) where f(x)=graph above, so do I just take the area between 0 and 2?

Does this seem ok?

I got my integral to be 6x-2e to the x/2 - e to the x)

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#11

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#11

(Original post by

Another unrelated question,https://gyazo.com/f7f4f586fbe793f267a91b46ceda23f9

am not sure what to do in this question.

I did integral between (f(x)-g(x) where f(x)=graph above, so do I just take the area between 0 and 2?

Does this seem ok?

I got my integral to be 6x-2e to the x/2 - e to the x)

**dwrfwrw**)Another unrelated question,https://gyazo.com/f7f4f586fbe793f267a91b46ceda23f9

am not sure what to do in this question.

I did integral between (f(x)-g(x) where f(x)=graph above, so do I just take the area between 0 and 2?

Does this seem ok?

I got my integral to be 6x-2e to the x/2 - e to the x)

Upload a pic of your working if you want more detailed help. Thanks.

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#12

(Original post by

Think you have the right idea, you want to integrate the difference of the curves from x = 0 up until the intersection point. The intersection point is not 2 however.

Upload a pic of your working if you want more detailed help. Thanks.

**mqb2766**)Think you have the right idea, you want to integrate the difference of the curves from x = 0 up until the intersection point. The intersection point is not 2 however.

Upload a pic of your working if you want more detailed help. Thanks.

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#13

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#13

(Original post by

I redone the factorising bit and let x= e to the x/2 and got a quadratic in x and still got x=-3 and x=2?

**dwrfwrw**)I redone the factorising bit and let x= e to the x/2 and got a quadratic in x and still got x=-3 and x=2?

e^2

and

6 - e^1

are different. I guess you've done a disguised quadratic and forgot to map back to the original x variable. I guess youve used x for both the original and transformed variables and got confused.

Last edited by mqb2766; 3 months ago

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#14

(Original post by

Can you pls try and upload your working in an image.

e^2

and

6 - e^1

are different. I guess you've done a disguised quadratic and forgot to map back to the original x variable. I guess youve used x for both the original and transformed variables and got confused.

**mqb2766**)Can you pls try and upload your working in an image.

e^2

and

6 - e^1

are different. I guess you've done a disguised quadratic and forgot to map back to the original x variable. I guess youve used x for both the original and transformed variables and got confused.

i got x=2In(2)

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#16

(Original post by

I presume youve checked this time, so just finish it off. Can you pls try and upload full working in an image in future. Its much easier to give help.

**mqb2766**)I presume youve checked this time, so just finish it off. Can you pls try and upload full working in an image in future. Its much easier to give help.

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#17

think i confused myself again.. ill send a pic of my work

(Original post by

I presume youve checked this time, so just finish it off. Can you pls try and upload full working in an image in future. Its much easier to give help.

**mqb2766**)

I presume youve checked this time, so just finish it off. Can you pls try and upload full working in an image in future. Its much easier to give help.

Last edited by dwrfwrw; 3 months ago

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#18

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#19

**mqb2766**)

I presume youve checked this time, so just finish it off. Can you pls try and upload full working in an image in future. Its much easier to give help.

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#20

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#20

Your integral is correct as are the limits. Youve just make some simple mistakes evaluating the definite integral when you sub the limits in. Just write it out more carefully. Note

e^ln(z) = z

as well and think about the form of the desired answer for the first ln() term.

e^ln(z) = z

as well and think about the form of the desired answer for the first ln() term.

Last edited by mqb2766; 3 months ago

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