Geometric sequences help

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

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Reply 1

mqb2766

Original post by dwrfwrw

1 = 2^?
2 = 2^?
...
64 = 2^?

(edited 2 years ago)

Reply 2

dwrfwrw

Original post by mqb2766

64 = 2^?

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

(edited 2 years ago)

Reply 3

mqb2766

Original post by 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...

You should know a doubling 2 sequence without taking logs?
So what power of 2 equals 64.

(edited 2 years ago)

Reply 4

dwrfwrw

Original post by mqb2766

You should know a doubling 2 sequence without taking logs?
So what power of 2 equals 64.

Yeah it's 6?
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..

(edited 2 years ago)

Reply 5

mqb2766

Original post by dwrfwrw

Yeah it's 6?

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

Reply 6

dwrfwrw

Original post by mqb2766

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

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

Reply 7

mqb2766

Original post by 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

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 so n=8.

What role does the - sign have?

(edited 2 years ago)

Reply 8

dwrfwrw

Original post by 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

Okay, thank you for your help I understood it now, thank you

Reply 9

dwrfwrw

Original post by mqb2766

1 = 2^?
2 = 2^?
...
64 = 2^?

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)

Reply 10

mqb2766

Original post by dwrfwrw

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.

Reply 11

dwrfwrw

Original post by mqb2766

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?

Reply 12

mqb2766

Original post by 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?

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.

(edited 2 years ago)

Reply 13

dwrfwrw

Original post by mqb2766

oh i figured it out yea i got confused,
i got x=2In(2)

Reply 14

mqb2766

Original post by dwrfwrw

oh i figured it out yea i got confused,
i got x=2In(2)

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.

Reply 15

dwrfwrw

Original post by 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.

sure only problem is that my phone is shared with my mom so it might be a bit awkward but ill definitely do that next time

Reply 16

dwrfwrw

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

Original post by 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.

(edited 2 years ago)

Original post by 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.

think i need help again aaa

Reply 19

mqb2766

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.