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, __, __, __, __, 963, __, __, __, __, 96?