Question #b09e5

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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)#