How do you find the geometric means in the sequence #3, , , , , 96#?

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Answer 1 · 2017-09-01T14:39:12

#6, 12, 24, and, 48.#

We will solve the Problem in #RR.#

Let #G_i, i=1" to "4,# be the desired #4# GMs. btwn. #3, and, 96.#

Clearly, then, #3, G_1, G_2, G_3, G_4, 96,# form a GP.

If, #t_n; n in NN# denotes the #n^(th)# term of the GP, then, we have,

#t_1=3, and, t_6=96," giving, "t_6/t_1=32.........(1).#

Following the Usual Notation of a GP,

#because t_n=a_1*r^(n-1) :. (1) rArr (a_1r^5)/a_1=32, or, r=32^(1/5)=2.#

#:. G_1=t_2=t_1*r=3*2=6, &," similarly, " #

#G_2=t_3=r*t_2=2*6=12, G_3=24, G_4=48.#

Thus, #6, 12, 24, and, 48# are the desired 4 GMs btwn.

#3, and, 96.#

Enjoy Maths.!

How do you find the geometric means in the sequence #3, __, __, __, __, 96#?