If n geometric means be inserted between a and b then {n^{th}} geometric mean will be

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8. Sequences and Series

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If $n$ geometric means be inserted between $a$ and $b$ then the ${n^{th}}$ geometric mean will be

A

$a\,{\left( {\frac{b}{a}} \right)^{\frac{n}{{n - 1}}}}$

B

$a\,{\left( {\frac{b}{a}} \right)^{\frac{{n - 1}}{n}}}$

C

$a\,{\left( {\frac{b}{a}} \right)^{\frac{n}{{n + 1}}}}$

D

$a\,{\left( {\frac{b}{a}} \right)^{\frac{1}{n}}}$

Solution

(c) If $n$ geometric means ${g_1},{g_2}…….{g_n}$ are to be inserted between two positive real numbers $a$ and $b$,

then $a,\;{g_1},\;{g_2}……{g_n},\;b$ are in $G.P. $Then

${g_1} = ar,\;{g_2} = a{r^2}……..{g_n} = a{r^n}$

So $b = a{r^{n + 1}} $

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

Now ${n^{th}}$ geometric mean $({g_n}) = a{r^n} = a\,\,{\left( {\frac{b}{a}} \right)^{n/(n + 1)}}$.

Aliter : As we have the ${m^{th}}$ $G.M.$ is given by

${G_m} = a{\left( {\frac{b}{a}} \right)^{\frac{m}{{n + 1}}}}$

Now replace $m$ by $n$ we get the required result.

Standard 11

Mathematics

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