# Question b09e5

Nov 24, 2017

 2^(8/5)*3^(1/5); 2^(27/10)*3^(2/5); 2^(19/5)*3^(3/5); 2^(49/10)*3^(4/5)#

#### Explanation:

We will solve the Problem in $\mathbb{R} .$

Suppose that, the reqd. $4$ GMs. btwn. $\sqrt{2} \mathmr{and} 192$ are,

${G}_{1} , {G}_{2} , {G}_{3} \mathmr{and} {G}_{4.}$

This would mean that, $\sqrt{2} , {G}_{1} , {G}_{2} , {G}_{3} , {G}_{4} , 192 , \ldots$ are in GP.

In other words, $192$ is the ${6}^{t h}$ term of the GP under

reference, of which $\sqrt{2}$ is the ${1}^{s t}$ term.

But, in the GP : $a , a r , a {r}^{2} , \ldots ,$ the ${n}^{t h}$ term ${t}_{n}$ is given by,

${t}_{n} = a \cdot {r}^{n - 1} , n \in \mathbb{N} .$

With $a = \sqrt{2} , n = 6 , {t}_{6} = 192 ,$ we have

$\sqrt{2} \cdot {r}^{5} = 192 \Rightarrow {r}^{5} = \frac{192}{\sqrt{2}} = \frac{{2}^{6} \cdot 3}{2} ^ \left(\frac{1}{2}\right) = {2}^{\frac{11}{2}} \cdot 3$

$\therefore r = {2}^{\frac{11}{10}} \cdot {3}^{\frac{1}{5}}$

$\therefore {G}_{1} = a r = {2}^{\frac{1}{2}} \cdot {2}^{\frac{11}{10}} \cdot {3}^{\frac{1}{5}} = {2}^{\frac{8}{5}} \cdot {3}^{\frac{1}{5}}$

${G}_{2} = {G}_{1} \cdot r = {2}^{\frac{8}{5}} \cdot {3}^{\frac{1}{5}} \cdot {2}^{\frac{11}{10}} \cdot {3}^{\frac{1}{5}} = {2}^{\frac{27}{10}} \cdot {3}^{\frac{2}{5}}$

${G}_{3} = {g}_{2} \cdot r = {2}^{\frac{27}{10}} \cdot {3}^{\frac{2}{5}} \cdot {2}^{\frac{11}{10}} \cdot {3}^{\frac{1}{5}} = {2}^{\frac{19}{5}} \cdot {3}^{\frac{3}{5}}$

${G}_{4} = {G}_{3} \cdot r = {2}^{\frac{19}{5}} \cdot {3}^{\frac{3}{5}} \cdot {2}^{\frac{11}{10}} \cdot {3}^{\frac{1}{5}} = {2}^{\frac{49}{10}} \cdot {3}^{\frac{4}{5}}$