Some general questions (1 Viewer)

[davidbarnes]

Got a maths test Wednesday and hust have some questions.

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As a power of 2 does
1
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256

equal 1
---
2^-8

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Tn = 24 x (-1/2)^n-1

With this geometric sequence, I have absolutely no idea how to solve this one. a = 24, and -1/2 = r, correct? How do I prove its goeometric (I know how to do that, just not for this one)

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With the below arithmetic sequence, how do I find the term of the sequence which is GREATER than 2500?

Tn = 6n - 42

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When using a graphics calcualtor to graph Y = 89 x 2^(5.06x)
How do I chose what to make X when using the graph mode on calcualtor, and not the statistics mode?

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A big thanks to any people who can help.

 

[davidbarnes]

just 2^-8

in form ar^n

5800 = 6n - 42 solve for n round up to nearest integer

We'd you get 5800 from in the last question? With No. 2, how does that prove its geotmetric?

 

[b0b444]

1. 1/256 = 2^-8

2. I think what 3unitz means is that the expression you gave, Tn = 24 x (-1/2)^n-1 , is in the from ar^(n-1) which is the form of a geometric sequence so it must be geometric. This is all true, but it sounds as though you knew that already. I don't know if this is really "proving" it. What I'd probably do is generate the first few terms of the sequence and show that the common ratio is the same.

3. I assume you mean for this to find the *first* term that is >2500. In this case 3unitz is exactly right.

Tn = 6n - 42 where Tn is greater than 2500.

therefore,

Tn = 6n - 42 = 2500

Now solve for n. Be careful to round the decimal correctly, n is an integer.

4. I'm sorry, I've never used a graphic calculator.