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Prove that the product of n geometric means between two quantities is equal to the nth power of a geometric mean of those two quantities.
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Let us suppose a and b are two numbers.
Let us say G is a number that is the Geometric mean of a and b
Therefore a,G and b must be in Geometric Progression or GP.
This means, common ratio = G/a = b/G
Or, G^{2} = ab
Or, G = ?(ab) ....................................... (1)
Now, let us say G_{1} , G_{2} , G3 ,.......Gn are n geomteric means between a and b.
Which means that
a , G_{1} , G_{2} , G_{3} ...... G_{n} , b form a G.P.
Note that the above GP has n+2 terms and the first term is a and last term is b, which
is also the (n+2)^{th} term
Hence, b = ar^{n+21}
where a is the first term.
So,
b = ar^{n+1}
r = (b/a)^{1/n+1 }..........................................(2)
Now the product of GP becomes
Product = G_{1}G_{2}G_{3}......G_{n}
_{ }
_{ }= (ar)(ar^{2})(ar^{3})..................(ar^{n})_{ }
= a^{n} × r^{(1+2+3+4+.............+n)}
= a^{n} × r^{n(1+n)/2}
Putting the value of r from equation 2 , we get
= a^{n} ×(b/a)^{n(1+n)/2(n+1)}
= (ab)^{n/2}
= (?ab)^{n}
Now, putting the value from equation 1 we get
Product = G^{n}
Or, G_{1}G_{2}G_{3}......G_{n }= G^{n}
Thank you
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