Answer :

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 geometric 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 the last term is b, which is also the (n+2)^{th} term

Hence, b = ar^{n+2–1}

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}

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