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Mathematics 16 Online
OpenStudy (anonymous):

a11=72 and a13=1/2 use the geometric mean to find the value of a12???

OpenStudy (anonymous):

do you know what a geometric mean is?

OpenStudy (anonymous):

the geometric mean of \(a\) and \(b\) is \[\sqrt{ab}\]

OpenStudy (anonymous):

I thought it was like the average # used during a pattern but idk not sure

OpenStudy (anonymous):

it is what i wrote above

OpenStudy (anonymous):

Ok I see it

OpenStudy (anonymous):

multiply the numbers, then take the square root

OpenStudy (anonymous):

a11,a12,a13 same as ar,ar2,ar3 and here a=72,ar2=12 so ar2a=r2=1272=1144 if r2=1144 then r=112

OpenStudy (anonymous):

no

OpenStudy (anonymous):

if you are using the geometric mean, then \[72\times \frac{1}{2}=36\] and \[\sqrt{36}=6\]

OpenStudy (anonymous):

or \[ar=72, ar^2,ar^3=\frac{1}{2}\implies r^2=\frac{ar^3}{ar}=\frac{1}{2\times 72}=\frac{1}{144}\] this tells you \(r=\frac{1}{12}\) and so \[72, 6, \frac{1}{2}\] is the sequence

OpenStudy (anonymous):

Idk I got 16

OpenStudy (anonymous):

Ill figure it out

OpenStudy (anonymous):

\[\large \sqrt{ab}=\sqrt{72\times \frac{1}{2}}=\sqrt{36}=6\]is the computation

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