# Geometric progression

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?

Those are both powers of 2.

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)

dextercioby
Homework Helper
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.

It turns out the answer is -4

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

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

matt grime
Homework Helper
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.

HallsofIvy
Homework Helper
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.

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

Daniel.

Since r= 256, a(2563)= 256 = 1/2562= 1/65536.

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 $a = r = \pm 4$.

HallsofIvy
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 $a = r = \pm 4$.