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Question: If \(n\) geometric means be inserted between \(a\) and \(b\)then the \(n ^ { t h }\) geometric m...

If nn geometric means be inserted between aa and bbthen the nthn ^ { t h } geometric mean will be.

A

a(ba)nn1a \left( \frac { b } { a } \right) ^ { \frac { n } { n - 1 } }

B

a(ba)n1na \left( \frac { b } { a } \right) ^ { \frac { n - 1 } { n } }

C

a(ba)nn+1a \left( \frac { b } { a } \right) ^ { \frac { n } { n + 1 } }

D

a(ba)1na \left( \frac { b } { a } \right) ^ { \frac { 1 } { n } }

Answer

a(ba)nn+1a \left( \frac { b } { a } \right) ^ { \frac { n } { n + 1 } }

Explanation

Solution

If nn geometric means g1,g2..gng _ { 1 } , g _ { 2 } \ldots \ldots . . g _ { n } are to be inserted between two positive real numbers aa and bb, then are in G.P. Then

So b=arn+1r=(ba)1/(n+1)b = a r ^ { n + 1 } \Rightarrow r = \left( \frac { b } { a } \right) ^ { 1 / ( n + 1 ) }

Now nthn ^ { t h } geometric mean (gn)=arn=a(ba)n/(n+1)\left( g _ { n } \right) = a r ^ { n } = a \left( \frac { b } { a } \right) ^ { n / ( n + 1 ) }.

Aliter : As we have the mthm ^ { t h } G.M. is given by

Gm=a(ba)mn+1G _ { m } = a \left( \frac { b } { a } \right) ^ { \frac { m } { n + 1 } }

Now replace mm by nn we get the required result.