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

Sep 1, 2017

$6 , 12 , 24 , \mathmr{and} , 48.$

#### Explanation:

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

Let ${G}_{i} , i = 1 \text{ to } 4 ,$ be the desired $4$ GMs. btwn. $3 , \mathmr{and} , 96.$

Clearly, then, $3 , {G}_{1} , {G}_{2} , {G}_{3} , {G}_{4} , 96 ,$ form a GP.

If, t_n; n in NN denotes the ${n}^{t h}$ term of the GP, then, we have,

${t}_{1} = 3 , \mathmr{and} , {t}_{6} = 96 , \text{ giving, } {t}_{6} / {t}_{1} = 32. \ldots \ldots . . \left(1\right) .$

Following the Usual Notation of a GP,

$\because {t}_{n} = {a}_{1} \cdot {r}^{n - 1} \therefore \left(1\right) \Rightarrow \frac{{a}_{1} {r}^{5}}{a} _ 1 = 32 , \mathmr{and} , r = {32}^{\frac{1}{5}} = 2.$

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

${G}_{2} = {t}_{3} = r \cdot {t}_{2} = 2 \cdot 6 = 12 , {G}_{3} = 24 , {G}_{4} = 48.$

Thus, $6 , 12 , 24 , \mathmr{and} , 48$ are the desired 4 GMs btwn.

$3 , \mathmr{and} , 96.$

Enjoy Maths.!