InterviewSolution
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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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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, G2 = ab Or, G = ?(ab)... (1) Now, let us say G1, G2, G3, .......Gn are n geometric means between a and b. Which means that a, G1, G2, G3 ...... Gn, 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 = arn+2–1 where a is the first term. So, b = arn+1 r = (b/a)1/n+1 ....(2) Now the product of GP becomes Product = G1G2G3......Gn = (ar)(ar2)(ar3).....(arn) = an r(1+2+3+4+.............+n) = an rn(1+n)/2 Putting the value of r from equation 2 , we get = an (b/a)n(1+n )/2(n+1) = (ab)n/2 = (?ab)n Now, putting the value from equation 1, we get, Product = Gn Or, G1G2G3......Gn = Gn |
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