How do you write a rule for the nth term of the geometric term given the two terms $a_3=25, a_6=-25/64$ ?

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How do you write a rule for the nth term of the geometric term given the two terms $a_3=25, a_6=-25/64$ ?

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Answer 1 · 2018-02-24T23:20:17

$400(-1/4)^(n-1)$

Explanation:

The nth term of a geometric series is given by:

$ar^(n-1)$

Where $bba$ is the first term, $bbr$ is the common ratio and $bbn$ is the nth term.

We have:

Third term $= 25$

Sixth term $= -25/64$

$:.$

$a_3=ar^(2)=25color(white)(888888.8)[1]$

$a_6=ar^(5)=-25/64color(white)(8888)[2]$

Solving for $bba$ and $bbr$

$[1]$

$a=25/r^2$

Substituting in $[2]$

$(25/r^2)r^5=-25/64$

Multiply by $r^2/25$

$r^5=-r^2/64$

$64r^5+r^2=0$

Factor out $bbr$

$r(64r^4+r)=0$

We know $bbr$ can't be zero.

$(64r^4+r)=0$

Factor out $bbr$ again.

$r(64^3+1)=0$

As before:

$64r^3+1=0$

$r^3=-1/64$

$r=root(3)(-1/64)=-1/4$

Using this in $[1]$

$a(-1/4)^2=25$

$a=25/(-1/4)^2=25/(1/64)=400$

So we have:

$400(-1/4)^(n-1)$ for our nth term.

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How do you write a rule for the nth term of the geometric term given the two terms $a_3=25, a_6=-25/64$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you write a rule for the nth term of the geometric term given the two terms a_3=25, a_6=-25/64a_3=25, a_6=-25/64?

1 Answer

Explanation:

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How do you write a rule for the nth term of the geometric term given the two terms $a_3=25, a_6=-25/64$ ?