# Geometric sequences help

#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?
0
3 months ago
#2
(Original post by 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?
1 = 2^?
2 = 2^?
...
64 = 2^?
Last edited by mqb2766; 3 months ago
0
#3
(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
Last edited by dwrfwrw; 3 months ago
0
3 months ago
#4
(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.
Last edited by mqb2766; 3 months ago
0
#5
(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..
Last edited by dwrfwrw; 3 months ago
0
3 months ago
#6
(Original post by dwrfwrw)
Yeah it's 6?
Of course and that is the n-1 term so n=7.
0
#7
(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
0
3 months ago
#8
(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?
Last edited by mqb2766; 3 months ago
0
#9
(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
0
#10
(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)
0
3 months ago
#11
(Original post by 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)
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.
0
#12
(Original post by 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.
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?
0
3 months ago
#13
(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?
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
0
#14
(Original post by mqb2766)
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.
oh i figured it out yea i got confused,
i got x=2In(2)
0
3 months ago
#15
(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.
0
#16
(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
0
#17
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.
Last edited by dwrfwrw; 3 months ago
0
#18
0
#19
(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
0
3 months ago
#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.
Last edited by mqb2766; 3 months ago
0
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