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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?

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- Thread starter rush
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- #1

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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?

- #2

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Those are both powers of 2.

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- #4

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Daniel.

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It turns out the answer is -4

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

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

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Actually, it's [itex]\pm 4[/itex]. See why?

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matt grime

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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.

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HallsofIvy

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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 ar^{3}= 256 and ar^{7}= 65536 so, dividing the second equation by the first, r^{7}/r^{3}= r^{4}= 65536/256= 256.

Since r= 256, a(256^{3})= 256 = 1/256^{2}= 1/65536.

In the example you started with, 4th term = 256 and the 8th term = 65536,

we know ar

Since r= 256, a(256

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Daniel.

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

you mean [itex]r^4 = 256 \Longrightarrow r = \pm 4[/itex] (over [itex]\mathbb{R}[/itex]) so

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

or just [itex]a = r = \pm 4[/itex].

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HallsofIvy

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Data said:you mean [itex]r^4 = 256 \Longrightarrow r = \pm 4[/itex] (over [itex]\mathbb{R}[/itex]) so

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

or just [itex]a = r = \pm 4[/itex].

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

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