Finding First Term in Geometric Progression with Given Terms?

[rush]

Hi there,
Can anybody help please.

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

 

[hypermorphism]

Those are both powers of 2.

 

[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)

 

[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.

 

[rush]

It turns out the answer is -4

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

 

[Data]

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

 

[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.

 

[HallsofIvy]

The general term of a geometric sequence is ar^n-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 ar^i-1= x. If you know "the jth term is y" then you know ar^j-1= y. Divide the second equation by the first and the "a"s cancel: r^j-1/r^i-1= r^j-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.

 

[dextercioby]

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

Daniel.

 

[Data]

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.

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