Chapter 9.4 - Geometric Mean

In the previous section, we saw the basic details about G.P. We saw some solved examples also. In this section, we will see two more solved examples. Later in this section, we will see geometric mean.

Solved example 9.15
Find the sum of the sequence 7, 77,777, 7777, . . . to n terms.
Solution:
1. We are asked to find the sum of the series: 7 + 77 + 777 + 7777 + . . . to n terms
2. This is not a geometric series. But we can relate it to a geometric series as follows:
$\begin{array}{ll}
{}&{S_n}
&{}={}& {7+77+777+7777+~.~.~\text{to n terms}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {7[1+11+111+1111+~.~.~\text{to n terms}]~\color{green}{\text{(Taking out common factor 7)}}}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {7 × \frac{9}{9}[1+11+111+1111+~.~.~\text{to n terms}]~\color{green}{\text{(Multiplying numerator and denominator by 9)}}}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {7 × \frac{1}{9}[9+99+999+9999+~.~.~\text{to n terms}]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {\frac{7}{9}[(10-1)+(100-1)+(1000-1)+(10000-1)+~.~.~\text{to n terms}]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {\frac{7}{9}[(10+100+1000+10000+~.~.~.~\text{to n terms})~-1-1-1-1-~.~.~\text{to n terms}]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {\frac{7}{9}[(10+100+1000+10000+~.~.~.~\text{to n terms})~-(1+1+1+1-~.~.~\text{to n terms})]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {\frac{7}{9} \left[\frac{10(10^n - 1)}{10-1}~-~n\right]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {\frac{7}{9} \left[\frac{10(10^n - 1)}{9}~-~n\right]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

Solved example 9.16
A person has 2 parents, 4 grand parents, 8 great grand parents and so on. Find the number of his ancestors during the 10 generations preceding his own.
Solution:
1. Fig.9.4 below shows the ancestors in order:

Fig.9.4

• The yellow square at the bottom represents the person under consideration.
• The cyan squares represent the parents. There are two parents.
• The green squares represent the grand parents. There are four grand parents.
• The magenta squares represent the great grand parents. There are eight great grand parents.
2. So the number of ancestors in each generation can be written as a sequence. We get: 2, 4, 8, 16, . . .
• This is a G.P with a = 2 and r = 2.
3. Sum of ten terms of this G.P will give the total number of ancestors in ten generations.
• We can write:
$S_{10}~=~\frac{a(r^n - 1)}{r-1}~=~\frac{2(2^{10} - 1)}{2-1}~=~\frac{2(2^{10} - 1)}{1}~=~2046$

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Geometric mean

This can be written in 5 steps:
1. Suppose that, we are given two numbers a and b
• We can find a number G in such a way that, a, G, b form a G.P
2. This G can be calculated in two simple steps:
(i) Since a, G, b is a G.P, we can write: $\frac{G}{a}~=~\frac{b}{G}$
(ii) From this we get: $G^2~=~ab$
⇒ $G=\sqrt{ab}$
3. This G is called the Geometric mean of a and b.
• Geometric mean is abbreviated as G.M
4. Let us see an example:
    ♦ Let the two numbers a and b be 2 and 8
    ♦ Then the G.M of 2 and 8 = $\sqrt{2 × 8}~=~\sqrt{16}$ = 4
5. We inserted just one number G between a and b. In fact we can insert as many numbers as we like between two numbers a and b so that the resulting sequence is a G.P.
• The following solved example will demonstrate the steps.

Solved example 9.17
Insert three numbers between 1 and 256 so that the resulting sequence is a G.P
Solution:
1. The resulting G.P will be in the form: 1, G₁, G₂, G₃, 256
    ♦ The first term of this G.P is 1
    ♦ The last term is 256
    ♦ Total number of terms = 5
2. So we can write:
256 = $a r^{n-1} ~=~1 × r^{5-1} ~=~r^4$
⇒ $256~=~4^4~=~r^4$
⇒ r = 4
3. Now we can write all the intermediate terms:
2^nd term G₁ = ar = 1 × 4 = 4
3^rd term G₂ = ar² = 1 × 4² = 16
4^th term G₃ = ar³ = 1 × 4³ = 64
4. So the resulting G.P is: 1, 4, 16, 64, 256
• The three numbers to be inserted are: 4, 16 and 64

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Relation between A.M and G.M

This can be written in 4 steps:
1. Consider two positive real numbers a and b
    ♦ Let A be the A.M of a and b
    ♦ Let G be the G.M of a and b
2. Then we can write: $A = \frac{a+b}{2}$
• Also we can write: $G=\sqrt{ab}$
3. Taking the difference, we get:
$A-G~=~ \frac{a+b}{2}~-~\sqrt{ab}~=~\frac{a+b-2 \sqrt{ab}}{2}~=~\frac{\left(\sqrt{a}- \sqrt{b} \right)^2}{2}$
• So we can write: $A-G~=~\frac{\left(\sqrt{a}- \sqrt{b} \right)^2}{2}$
4. Consider the R.H.S of the above expression. This R.H.S cannot be less than zero.
• So we can write:
    ♦ G will be either equal to A or less than A
    ♦ G can never be greater than A

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The link below gives some more solved examples

Exercise 9.3

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In the next section we will see special series.

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