Question

Let five geometric means are inserted between $ \dfrac{{32}}{3} $ and \[\dfrac{{243}}{2}\] then find the sum of all the geometric means.

Answer 1

Hint: To solve this problem, i.e., to find sum of all the five geometric means, we will first consider the general sequence of a geometric mean, here we will take the given value as, the first term equals to $ \dfrac{{32}}{3} $ while the last term equals to \[\dfrac{{243}}{2}\], then by dividing these values, we will get the value of r, i.e., common ratio, now after that on applying the formula of sum of geometric mean we will get our required answer.

Complete step-by-step answer:
We have been given that five geometric means are to be inserted between $ \dfrac{{32}}{3} $ and \[\dfrac{{243}}{2}\], so we need to first find five GM and then the sum of all the geometric means, which we will find first.
We know that, the general sequence of GM is, \[a,{\text{ }}ar,{\text{ }}a{r^2},{\text{ }}a{r^3},{\text{ }}a{r^4},{\text{ }}a{r^5},{\text{ }}a{r^6} \ldots .\]
Here, according to the question, $ a = \dfrac{{32}}{3} $ and \[a{r^6} = \dfrac{{243}}{2}.\]
Now, to find the value of r, i.e., the common ratio, we will divide \[a{r^6}\] by \[a,\] we get
 $
\dfrac{{a{r^6}}}{a} = \dfrac{{243/2}}{{32/3}} \\
{r^6} = \dfrac{{729}}{{64}} \\
{r^6} = \dfrac{{{3^6}}}{{{2^6}}} \\
\Rightarrow r = \dfrac{3}{2} \\
  $
Thus, \[a{\text{ }} = {\text{ }}\dfrac{{32}}{3},ar = \left( {\dfrac{{32}}{3}} \right)\left( {\dfrac{3}{2}} \right) = 16,a{r^2} = \left( {\dfrac{{32}}{3}} \right)\left( {\dfrac{3}{2}} \right)\left( {\dfrac{3}{2}} \right) = 24,a{r^3} = \left( {\dfrac{{32}}{3}} \right)\left( {\dfrac{3}{2}} \right)\left( {\dfrac{3}{2}} \right)\left( {\dfrac{3}{2}} \right) = 36,\]
\[a{r^4} = \left( {\dfrac{{32}}{3}} \right)\left( {\dfrac{3}{2}} \right)\left( {\dfrac{3}{2}} \right)\left( {\dfrac{3}{2}} \right)\left( {\dfrac{3}{2}} \right) = 54,a{r^5} = \left( {\dfrac{{32}}{3}} \right)\left( {\dfrac{3}{2}} \right)\left( {\dfrac{3}{2}} \right)\left( {\dfrac{3}{2}} \right)\left( {\dfrac{3}{2}} \right)\left( {\dfrac{3}{2}} \right) = 81,a{r^6} = \dfrac{{243}}{2}\]
So, the GP of geometric mean is, $ \dfrac{{32}}{3},16,24,36,54,81,\dfrac{{243}}{2} $
Now, to find the sum of geometric mean, we use the formula mentioned below.
Sum $ = a\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right) $
On putting the values of $ a = 16,r = \dfrac{3}{2} $ and $ n = 5, $ in the above formula, we get
Sum of geometric mean $ = 16\left( {\dfrac{{{{(3/2)}^5} - 1}}{{3/2 - 1}}} \right) $
 $
= 16\left( {\dfrac{{{{(3/2)}^5} - 1}}{{1/2}}} \right) \\
= 16\left( {\dfrac{{(243/32) - 1}}{{1/2}}} \right) \\
= 32\left( {\dfrac{{211}}{{32}}} \right) \\
= 211 \\
  $
Thus, the sum of all the geometric means is \[211.\]

Note: Students should note that in the solution, we have taken once, \[a = \dfrac{{32}}{3},\] while evaluating the sum of the geometric means, we have taken \[a = 16,\] because, we are asked only about the geometric means of five terms. Thus, we have taken \[\;a = 16.\]