Question

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

Answer 1

Hint: In the above problem the first term ${a_1}$and the sixth term ${a_6}$ is given. By making use of this we need to find the remaining terms of the geometric sequence. A geometric sequence is also called a geometric progression. It is actually a sequence of non-zero numbers such that there in the sequence each term subsequent to the first term goes to the next by always multiplying by the same fixed non-zero number called the common ratio `r’.
 Here we first need to find the common ratio r and using this common ratio we find the remaining terms of the sequence.
 Here we make use of the formula, ${a_n} = a{r^{n - 1}}$, where a is the first term and r is the common ratio.
In the above problem we put $n = 2,3,4,5$.

Complete step by step solution:
Given ${a_1} = 3,$ ${a_6} = 96$
We need to find the remaining terms of the geometric sequence.
Generally we represent an geometric sequence by $a,ar,a{r^2},a{r^4},a{r^5},a{r^6}.....$
where ${a_1} = a,$ ${a_2} = ar,$ ${a_3} = a{r^2},$ ${a_4} = a{r^3}$ and so on.
Where, $a = $First term of geometric progression
$r = $ Common ratio
We now use the formula, ${a_n} = a{r^{n - 1}}$ ……(1)
Where $a = $first term of the geometric sequence which is 3.
We now make use of the formula given in the equation (1) for $n = 6$.
 ${a_6} = a{r^{6 - 1}}$
We know that ${a_6} = 96$ and $a = 3$.
$ \Rightarrow 96 = 3{r^5}$
Taking 3 to the R.H.S. and simplifying we get the required value of the common ratio r.
$ \Rightarrow \dfrac{{96}}{3} = {r^5}$
$ \Rightarrow 32 = {r^5}$.
We can easily see that 32 comes when 2 is multiplied five times.
i.e. $2 \times 2 \times 2 \times 2 \times 2 = {2^5} = 32$.
Comparing this with ${r^5} = 32$, we get the common ratio.
$ \Rightarrow {r^5} = {2^5}$
$ \Rightarrow r = 2$.
To obtain the second term put $n = 2$ in the equation (1), we get,
${a_2} = a{r^{2 - 1}}$
$ \Rightarrow {a_2} = ar$
Substituting $a = 3$ and $r = 2$, we get,
$ \Rightarrow {a_2} = 3 \times 2$
$ \Rightarrow {a_2} = 6$.
To obtain the third term put $n = 3$ in the equation (1), we get,
${a_3} = a{r^{3 - 1}}$
$ \Rightarrow {a_3} = a{r^2}$
Substituting $a = 3$ and $r = 2$, we get,
$ \Rightarrow {a_3} = 3 \times {2^2}$
$ \Rightarrow {a_3} = 3 \times 4$
$ \Rightarrow {a_3} = 12$.
To obtain the fourth term put $n = 4$ in the equation (1), we get,
${a_4} = a{r^{4 - 1}}$
$ \Rightarrow {a_4} = a{r^3}$
Substituting $a = 3$ and $r = 2$, we get,
$ \Rightarrow {a_4} = 3 \times {2^3}$
$ \Rightarrow {a_4} = 3 \times 8$
$ \Rightarrow {a_4} = 24$.
To obtain the fifth term put $n = 5$ in the equation (1), we get,
${a_5} = a{r^{5 - 1}}$
$ \Rightarrow {a_5} = a{r^4}$
Substituting $a = 3$ and $r = 2$, we get,
$ \Rightarrow {a_5} = 3 \times {2^4}$
$ \Rightarrow {a_5} = 3 \times 16$
$ \Rightarrow {a_5} = 48$.
$\therefore {a_1} = 3,$ ${a_2} = 6,$ ${a_3} = 12,$ ${a_4} = 24,$ ${a_5} = 48,$ ${a_6} = 96.$
Hence $6,12,24,48$ are the desired four geometric means between 3 and 96.

Note:
If the same number is not multiplied to each number in the sequence, then we cannot form the geometric sequence.
We must remember the formula to find the nth term of an geometric progression which is given by ${a_n} = a{r^{n - 1}}$
Where, ${a_n} = $nth term of geometric sequence.
$a = $first term of the geometric sequence.
$r = $common ratio
If we have given a list of numbers and asked to examine whether it is a geometric progression or not, we need to just find the common ratio r between the consecutive numbers. If the common ratio r comes to be the same, then the given list of numbers is a geometric progression. Otherwise it is not a geometric progression.