# Geometric progression

1. Apr 7, 2005

### rush

Hi there,

How can i find the first term in a geometric progression if i know that the 4th term = 256 and the 8th term = 65536?

2. Apr 7, 2005

### hypermorphism

Those are both powers of 2.

3. Apr 7, 2005

### rush

Sorry, perhaps that wasn't a good example, in general how do i find the 1st term (say a) in a geometric progression if i know 2 other other terms (say b and c)

4. Apr 7, 2005

### dextercioby

Use logarithmation.If 65536 is the 8-th term,then the first is $\log_{8} 65536$...(the first is just the ratio.If it's not,then u need two terms...)

Daniel.

5. Apr 7, 2005

### rush

It turns out the answer is -4

There must be a formula to work this out without using logarithmation

6. Apr 7, 2005

### Data

Actually, it's $\pm 4$. See why?

7. Apr 7, 2005

### matt grime

What is the definition of the n'th term of a GP with initial term a and ratio r?

ar^{n-1}, right?

so given two terms you've two unknowns and fortunately you can solve for them, though at some point you will need to take some roots.

8. Apr 7, 2005

### HallsofIvy

Staff Emeritus
The general term of a geometric sequence is arn-1 where a is the first value (n= 1) and r is the "common ratio". If you know "the i th term is x" then you know ari-1= x. If you know "the jth term is y" then you know arj-1= y. Divide the second equation by the first and the "a"s cancel: rj-1/ri-1= rj-i= y/x. Now it's not really necessary to use logarithms to find r- just use the "j-i" root. Once you know r, you can solve either of the original equations for a.

In the example you started with, 4th term = 256 and the 8th term = 65536,
we know ar3= 256 and ar7= 65536 so, dividing the second equation by the first, r7/r3= r4= 65536/256= 256.
Since r= 256, a(2563)= 256 = 1/2562= 1/65536.

Last edited: Apr 7, 2005
9. Apr 7, 2005

### dextercioby

Halls,$r^{4}=256$ has 4 solutions...$\left\{\pm 4,\pm 4i\right\}$...

Daniel.

10. Apr 7, 2005

### Data

you mean $r^4 = 256 \Longrightarrow r = \pm 4$ (over $\mathbb{R}$) so

$$a(\pm 4)^3 = 256 \Longrightarrow a =\biggr \{ \begin{array}{cc} 4 & \mbox{if} \ r=4 \\ -4 & \mbox{if} \ r = -4\end{array}$$

or just $a = r = \pm 4$.

11. Apr 7, 2005

### HallsofIvy

Staff Emeritus
Oops. My math is fine- but my arithmetic is terrible!