Question

If n geometric means be inserted between a and b then the ${n^{th}}$ geometric mean will be
1. $a{\left( {\dfrac{b}{a}} \right)^{\dfrac{n}{{\left( {n - 1} \right)}}}}$
2. $a{\left( {\dfrac{b}{a}} \right)^{\dfrac{{\left( {n - 1} \right)}}{n}}}$
3. $a{\left( {\dfrac{b}{a}} \right)^{\dfrac{n}{{\left( {n + 1} \right)}}}}$
4. $a{\left( {\dfrac{b}{a}} \right)^{\dfrac{1}{n}}}$

Answer 1

Hint: Here it is given that n geometric means are inserted between a and b and we need to calculate the value of the ${n^{th}}$ geometric mean. We need to frame a sequence according to the given information. Then, we need to count the number of terms present in that sequence. Then, we shall obtain the common ratio by applying the formula. Finally, we need to apply it in the ${n^{th}}$ geometric mean to obtain the desired answer.
Formula to be used:
${n^{th}}$ term of the sequence $ = a{r^{n - 1}}$
Where a is the first term and r is the common ratio.

Complete step-by-step answer:
It is given that n geometric means are inserted between a and b. We need to calculate the value of the ${n^{th}}$ geometric mean.
Since we are given that n geometric means are inserted between a and b, we can write it mathematically as $\left( {a,{G_1},{G_2},...,{G_n},b} \right)$ where ${G_1}$ is the first geometric mean, ${G_2}$ is the second geometric mean and ${G_n}$ is the ${n^{th}}$ geometric mean.
In this sequence $\left( {a,{G_1},{G_2},...,{G_n},b} \right)$, we can able to decide that there will be $n + 2$ terms.
That is, $n + 2$are presented in the sequence $\left( {a,{G_1},{G_2},...,{G_n},b} \right)$
Also, we can say that a is the first term and b is the last term.
Let us assume that r is the common ratio of the geometric sequence $\left( {a,{G_1},{G_2},...,{G_n},b} \right)$
We know that the formula to find the ${n^{th}}$term of the sequence is $a{r^{n - 1}}$ when n terms are presented.
Here, we have $n + 2$terms.
Thus, we have $b = a{r^{\left( {n + 2} \right) - 1}}$
$ \Rightarrow b = a{r^{n + 1}}$
Now, we shall calculate the value of r.
$ \Rightarrow \dfrac{b}{a} = {r^{n + 1}}$
Hence, we get \[r = {\left( {\dfrac{b}{a}} \right)^{\dfrac{1}{{n + 1}}}}\]
We are asked to calculate the ${n^{th}}$ geometric mean in the sequence $\left( {a,{G_1},{G_2},...,{G_n},b} \right)$
Thus, we have $n + 1$ terms.
We know that the formula to find the ${n^{th}}$term of the sequence is $a{r^{n - 1}}$ when n terms are presented.
Hence, ${n^{th}}$ geometric mean $ = a{r^{\left( {n + 1} \right) - 1}}$
So, ${n^{th}}$ geometric mean $ = a{r^n}$
We need to apply \[r = {\left( {\dfrac{b}{a}} \right)^{\dfrac{1}{{n + 1}}}}\] in the above equation.
That is, ${n^{th}}$ geometric mean $ = a{\left( {\dfrac{b}{a}} \right)^{\dfrac{n}{{n + 1}}}}$
Therefore, we got the required ${n^{th}}$ geometric mean, and option 3) is the answer.

So, the correct answer is “Option 3”.

Note: Before getting into the solution to obtain the ${n^{th}}$ geometric mean, we need to calculate the value of the last term. Here the last term is b. So, we need to calculate the value of b so that we can get the required common ratio. Without getting the value of the common ratio, we cannot find the desired ${n^{th}}$ geometric mean.