Question #b09e5

Question

1397 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-24T06:10:43

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

We will solve the Problem in $RR.$

Suppose that, the reqd. $4$ GMs. btwn. $sqrt2 and 192$ are,

$G_1,G_2,G_3 and G_4.$

This would mean that, $sqrt2,G_1,G_2,G_3,G_4,192,...$ are in GP.

In other words, $192$ is the $6^(th)$ term of the GP under

reference, of which $sqrt2$ is the $1^(st)$ term.

But, in the GP : $a,ar,ar^2,...,$ the $n^(th)$ term $t_n$ is given by,

$t_n=a*r^(n-1), n in NN.$

With $a=sqrt2, n=6, t_6=192,$ we have

$sqrt2*r^5=192 rArr r^5=192/sqrt2=(2^6*3)/2^(1/2)=2^(11/2)*3$

$:. r=2^(11/10)*3^(1/5)$

$:. G_1=ar=2^(1/2)*2^(11/10)*3^(1/5)=2^(8/5)*3^(1/5)$

$G_2=G_1*r=2^(8/5)*3^(1/5)*2^(11/10)*3^(1/5)=2^(27/10)*3^(2/5)$

$G_3=g_2*r=2^(27/10)*3^(2/5)*2^(11/10)*3^(1/5)=2^(19/5)*3^(3/5)$

$G_4=G_3*r=2^(19/5)*3^(3/5)*2^(11/10)*3^(1/5)=2^(49/10)*3^(4/5)$