A.1 Infinite Geometric Series

In the previous section, we saw infinite series related to binomial expansions. In this section, we will see infinite series related to geometric series.

• In chapter 9, we saw the general form of a geometric series. Let us write it again:
$a + ar + ar^2 + ar^3 ~+ ~.~.~.~+~ar^{n-1}$
• This is a finite series. There are n terms.
• We know that, for such a finite series, the sum is given by:
$S_n = \frac{a \left(1 - r^n \right)}{1-r}$

---

• Now let us suppose that, there are infinite number of terms. Then it will be an infinite geometric series.
• For such series also, we can derive a formula to find the sum.
• First we will see an example to learn the general features of infinite geometric series. It can be written in 7 steps:

1. Consider the geometric series:
$1 + \frac{2}{3} + \frac{4}{9} + ~.~.~.$
2. We see that, the first term, a = 1
3. Let us calculate r:
• $\frac{2}{3} \div 1 = \frac{2}{3}$

• $\frac{4}{9} \div \frac{2}{3} = \frac{4}{9} \times\frac{3}{2} = \frac{2}{3}$

• So we can write: $r = \frac{2}{3}$

4. Now we can write the sum of n terms:

$\begin{array}{ll}{}    &{S_n}    & {~=~}    &{\frac{a \left(1 - r^n \right)}{1-r}}    &{} \\
{}    &{}    & {~=~}    &{\frac{1 \times \left[1 - \left(\frac{2}{3} \right)^n \right]}{1-\left(\frac{2}{3} \right)}}    &{} \\
{}    &{}    & {~=~}    &{\frac{\left[1 - \left(\frac{2}{3} \right)^n \right]}{\frac{1}{3}}}    &{} \\
{}    &{}    & {~=~}    &{3 \left[1 - \left(\frac{2}{3} \right)^n \right]}    &{} \\
\end{array}$

5. We know that, when the number of terms in the series increases, n increases.
• When n increases, what happens to the term $\left(\frac{2}{3} \right)^n$ ?
◼ Let us put some convenient values for n:
(calculations can be easily done using a spreadsheet program)
• When n = 1, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^1 = 0.666666666666667$
• When n = 5, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^5 = 0.131687242798354$
• When n = 10, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^{10} = 0.017341529915833$
• When n = 20, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^{20} = 0.000300728659822$
• When n = 30, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^{30} = 0.00000521509505084655$

6. It is clear that, when n increases, $\left(\frac{2}{3} \right)^n$ decreases.
• When n approaches infinity, $\left(\frac{2}{3} \right)^n$ will approach zero. There will be more than a million zeros just to the right of the decimal point. For all practical purposes, it can be considered as zero.
7. Now consider the result in (4).
• Using that result, we can write:
$S_n = 3[1 - 0] = 3~\text{where n is infinity}$

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Let us see if we can write a general form of the above result. It can be done in 4 steps:
1. Consider the geometric series:
$a + ar + ar^2 + ar^3 ~+ ~.~.~.~+~ar^{n-1}~+ ~.~.~.$
• The common ratio r is a proper fraction. It can be +ve or -ve.
• Mathematically, we write this as: |r| < 1
(Proper fraction is a fraction in which denominator is larger than the numerator)
• We want the sum when n is infinity.
2. We have:
$S_n = \frac{a \left(1 - r^n \right)}{1-r}$
• This can be written as:
$S_n = \frac{a}{1-r}~-~\frac{a r^n}{1-r}$
3. Now we note an important point:
    ♦ Since |r| < 1,
    ♦ $r^n$ will approach zero
    ♦ when n approach infinity.
• So the result in (2) will become:

$\begin{array}{ll}{}    &{S_n}    & {~=~}    &{\frac{a}{1-r}~-~\frac{a r^n}{1-r}}    &{} \\
{}    &{}    & {~=~}    &{\frac{a}{1-r}~-~\frac{a \times 0}{1-r}}    &{} \\
{}    &{}    & {~=~}    &{\frac{a}{1-r}~-~0}    &{} \\
{}    &{}    & {~=~}    &{\frac{a}{1-r}}    &{} \\
\end{array}$

4. Symbolically, when sum of an infinite geometric series is written, we use "S" instead of "S_n".
• So we have: $S = \frac{a}{1-r}$

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Now we will see a solved example:

Solved example A.2
Find the sum to infinity of the following geometric progressions:
(i) $1, \frac{1}{2}, \frac{1}{2^2}, \frac{1}{2^3},~.~.~.$
(ii) $1, \frac{-1}{2}, \frac{1}{2^2}, \frac{-1}{2^3},  ~.~.~.$
(iii) $\frac{-5}{4}, \frac{5}{16}, \frac{-5}{64}, ~.~.~.$
Solution:
Part (i):
1. The infinite geometric series associated with the given G.P is:
$1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + ~.~.~.$
2. First term a is 1 and the common ratio r is $\frac{1}{2}$.
3. Absolute value of r is less than 1. So we can use the formula: $S = \frac{a}{1-r}$

$\begin{array}{ll}{}    &{S}    & {~=~}    &{\frac{a}{1-r}}    &{} \\
{}    &{}    & {~=~}    &{\frac{1}{1-\frac{1}{2}}}    &{} \\
{}    &{}    & {~=~}    &{2}    &{} \\
\end{array}$

Part (ii):
1. The infinite geometric series associated with the given G.P is:
$1 - \frac{1}{2} + \frac{1}{2^2} - \frac{1}{2^3} + ~.~.~.$
2. First term a is 1 and the common ratio r is $-\frac{1}{2}$.
3. Absolute value of r is less than 1. So we can use the formula: $S = \frac{a}{1-r}$

$\begin{array}{ll}{}    &{S}    & {~=~}    &{\frac{a}{1-r}}    &{} \\
{}    &{}    & {~=~}    &{\frac{1}{1- \left(-\frac{1}{2} \right)}}    &{} \\
{}    &{}    & {~=~}    &{\frac{1}{1+ \frac{1}{2}}}    &{} \\
{}    &{}    & {~=~}    &{\frac{1}{\frac{3}{2}}}    &{} \\
{}    &{}    & {~=~}    &{\frac{2}{3}}    &{} \\
\end{array}$

Part (iii):
1. The infinite geometric series associated with the given G.P is:
$\frac{-5}{4} + \frac{5}{16} - \frac{5}{64} + ~.~.~.$
2. First term a is $\frac{-5}{4}$ and the common ratio r is $-\frac{1}{4}$.
3. Absolute value of r is less than 1. So we can use the formula: $S = \frac{a}{1-r}$

$\begin{array}{ll}{}    &{S}    & {~=~}    &{\frac{a}{1-r}}    &{} \\
{}    &{}    & {~=~}    &{\frac{\frac{-5}{4}}{1- \left(-\frac{1}{4} \right)}}    &{} \\
{}    &{}    & {~=~}    &{\frac{\frac{-5}{4}}{1 + \frac{1}{4}}}    &{} \\
{}    &{}    & {~=~}    &{\frac{\frac{-5}{4}}{\frac{5}{4}}}    &{} \\
{}    &{}    & {~=~}    &{-1}    &{} \\
\end{array}$

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In the next section, we will see exponential series.

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Chapter 9.7 - Miscellaneous Examples on chapter 9

In the previous section, we completed a discussion on sequences and series. In this section, we will see some miscellaneous examples.

Solved example 9.21
If p^th, q^th, r^th and s^th terms of an A.P are in G.P, then show that (p-q), (q-r), (r-s) are also in G.P.
Solution:
1. We want to prove that (p-q), (q-r), (r-s) are in G.P.
• That means, we have to prove that $\frac{q-r}{p-q}~=~\frac{r-s}{q-r}$

2. Let 'a' be the first term and 'd' the common difference of the A.P.
• Then the given terms can be written as:
    ♦ p^th term = a_p = a + (p-1)d
    ♦ q^th term = a_q = a + (q-1)d
    ♦ r^th term = a_r = a + (r-1)d
    ♦ s^th term = a_s = a + (s-1)d
3. Given that, the above four terms are in G.P. So we can write:

$\begin{array}{ll}
{}&{\frac{a_q}{a_p}}
&{~=~\frac{a_r}{a_q}}& {~=~\frac{a_s}{a_r}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{a+(q-1)d}{a+(p-1)d}}
&{~=~\frac{a+(r-1)d}{a+(q-1)d}}& {~=~\frac{a+(s-1)d}{a+(r-1)d}}
&{\color {green} {\text{- - - - (a)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{a+(q-1)d~-~[a+(p-1)d]}{a+(p-1)d}}
&{~=~\frac{a+(r-1)d~-~[a+(q-1)d]}{a+(q-1)d}}& {~=~\frac{a+(s-1)d~-~[a+(r-1)d]}{a+(r-1)d}}
&{\color {green} {\text{- - - - (b)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{a+qd-d-a-pd+d}{a+(p-1)d}}
&{~=~\frac{a+rd-d-a-qd+d}{a+(q-1)d}}& {~=~\frac{a+sd-d-a-rd+d}{a+(r-1)d}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{(q-p)d}{a+(p-1)d}}
&{~=~\frac{(r-q)d}{a+(q-1)d}}& {~=~\frac{(s-r)d}{a+(r-1)d}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
(i) Line marked as (a):
This line is obtained by substituting the results from step (2)
(ii) Line marked as (b):
Here we apply componendo and dividendo rule:

$\begin{array}{ll}
{\text{If}}&{\frac{a}{b}~=~\frac{c}{d}}
&{\text{Then}}& {\frac{a-b}{b}~=~\frac{c-d}{d}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\text{Example:}}&{\frac{5}{2}~=~\frac{10}{4}}
&{\text{Then}}& {\frac{5-2}{2}~=~\frac{10-4}{4}~=~\frac{6}{4}~=~\frac{3}{2}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

4. Consider the above result in (3).
Take out the first two items. We get:

$\begin{array}{ll}
{}&{\frac{(q-p)d}{a+(p-1)d}}
&{~=~\frac{(r-q)d}{a+(q-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{p-q}{a+(p-1)d}}
&{~=~\frac{q-r}{a+(q-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{q-r}{p-q}}
&{~=~\frac{a+(q-1)d}{a+(p-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$ 

5. Consider again the result in (3).
Take out the last two items. We get:

$\begin{array}{ll}
{}&{\frac{(r-q)d}{a+(q-1)d}}
&{~=~\frac{(s-r)d}{a+(r-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{q-r}{a+(q-1)d}}
&{~=~\frac{r-s}{a+(r-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{r-s}{q-r}}
&{~=~\frac{a+(r-1)d}{a+(q-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

• From the line marked as (a) in step (3), we get:
$\frac{a+(r-1)d}{a+(q-1)d}~=~\frac{a+(q-1)d}{a+(p-1)d}$

• So we can write: $\frac{r-s}{q-r}~=~\frac{a+(q-1)d}{a+(p-1)d}$

6. Comparing the results in (4) and (5), we get:

$\frac{q-r}{p-q}~=~\frac{a+(q-1)d}{a+(p-1)d}~=~\frac{r-s}{q-r}$

• Hence the statement in (1) is proved.

Solved example 9.22
If a, b, c are in G.P and $a^{\frac{1}{x}}~=~b^{\frac{1}{y}}~=~c^{\frac{1}{z}}$, prove that x, y, z are in A.P.
Solution:
1. We want to prove that x, y, z are in A.P.
• That means, we have to prove that y - x = z - y
2. Given that $a^{\frac{1}{x}}~=~b^{\frac{1}{y}}~=~c^{\frac{1}{z}}$
Let us assume that, the three quantities are equal to k. So we can write:
$a^{\frac{1}{x}}~=~b^{\frac{1}{y}}~=~c^{\frac{1}{z}}~=~k$
3. Consider the above result in (2). Taking the first and last items, we get:

$\begin{array}{ll}
{}&{a^{\frac{1}{x}}}
&{~=~k}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\left(a^{\frac{1}{x}} \right)^x}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\left(a^{\frac{x}{x}} \right)}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\left(a^{1} \right)}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{a}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ In a similar way, we will get:
b = k^y and c = k^z

4. Now we get:
$\frac{b}{a}~=~\frac{k^y}{k^x}~=~k^{y-x}$
5. Also we get:
$\frac{c}{b}~=~\frac{k^z}{k^y}~=~k^{z-y}$
6. Given that a, b and c are in G.P. So we can write: $\frac{b}{a}~=~\frac{c}{b}$
7. So we can equate the results in (4) and (5). We get:
$\frac{b}{a}~=~\frac{c}{b}~=~k^{y-x}~=~k^{z-y}$
• Equating the powers of k, we get: y-x = z-y
• Hence the statement in (1) is proved.

Solved example 9.23
If a, b, c, d and p are different real numbers such that
$\left(a^2 + b^2 + c^2 \right)p^2~-~2(ab+bc+cd)p~+~\left(b^2 + c^2 + d^2 \right) ~\le~0$,
then show that a, b, c and d are in G.P.
Solution:
1. We want to prove that a, b, c and d are in G.P.
• That means, we have to prove that $\frac{b}{a}~=~\frac{c}{b}~=~\frac{d}{c}$.
2. The given inequality can be rearranged as follows:

$\begin{array}{ll}
{}&{\left(a^2 + b^2 + c^2 \right)p^2~-~2(ab+bc+cd)p~+~\left(b^2 + c^2 + d^2 \right)}
&{~\le~0}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\left(a^2 p^2 - 2abp + b^2 \right)~+~\left(b^2 p^2 - 2bcp + c^2 \right)~+~\left(c^2 p^2 - 2cdp + d^2 \right)}
&{~\le~0}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{(ap-b)^2~+~(bp-c)^2~+~(cp-d)^2}
&{~\ge~0}& {\color {green} {\text{- - - - (a)}}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
Line marked as (a):
• This line is a sum of three terms. Each of the three terms is a square.
    ♦ Given that: a, b, c, d and p are real numbers.
    ♦ So (ap-b), (bp-c) and (cp-d) will also be real numbers.
    ♦ Square of real numbers will be zero or +ve.
    ♦ So the sum of the three terms will be zero or +ve.
    ♦ Thus the sign in this line is changed to ≥.

3. The given inequality is only rearranged in the above step (2).
• The rearranged result in (2) is same as the given inequality.
• But the given inequality is ≤. In step (2), it changed to ≥.
• If both ≤ and ≥ are to be satisfied, the only possibility is that, both the expressions are equal to zero. 
4. So we can write: $(ap-b)^2~+~(bp-c)^2~+~(cp-d)^2~=~0$
• If the sum of certain squares is zero, each term in that sum must be zero.
• So we get:
(i) (ap-b)² = 0
(ii) (bp-c)² = 0
(iii) (cp-d)² = 0
5. If square of a real number is zero, then that real number must be zero.
So we get:
(i) ap-b = 0
(ii) bp-c = 0
(iii) cp-d = 0
• From 5(i), we get: $p=\frac{b}{a}$
• From 5(ii), we get: $p=\frac{c}{b}$
• From 5(iiii), we get: $p=\frac{d}{c}$
6. Based on the above step (5), we can write:
$\frac{b}{a}~=~\frac{c}{b}~=~\frac{d}{c}~=~p$
• Hence the statement in (1) is proved.

Solved example 9.24
If p, q, r are in G.P and the equations px² + 2qx + r = 0 and dx² + 2ex + f = 0 have a common root, then show that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in A.P.
Solution:
1. We want to prove that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in A.P.
• That means, we have to prove that:

$\begin{array}{ll}
{}&{\frac{e}{q} - \frac{d}{p}}
&{~=~\frac{f}{r} - \frac{e}{q}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{e}{q} + \frac{e}{q}}
&{~=~\frac{f}{r} + \frac{d}{p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{2e}{q}}
&{~=~\frac{f}{r} + \frac{d}{p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

2. Given that, p, q, r are in G.P. So we can write: q² = pr
3. Also given that, px² + 2qx + r = 0 and dx² + 2ex + f = 0 have a common root.
• Let us write the roots of the first equation. We get:

$\begin{array}{ll}
{}&{\frac{-2q \pm \sqrt{(2q)^2 ~-~4pr}}{2p}}
&{~=~\frac{-2q \pm \sqrt{4q^2 ~-~4pr}}{2p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{~=~}&{\frac{-2q \pm \sqrt{4pr ~-~4pr}}{2p}}
&{~=~\frac{-2q \pm \sqrt{0}}{2p}}& {\color {green} {\text{- - - - (a)}}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{~=~}&{\frac{-2q}{2p}}
&{~=~\frac{-q}{p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
Line marked as (a): From (2), we have: q² = pr

4. So the first equation has only one root, which is: $\frac{-q}{p}$
• Given that, the two equations have a common root. So $\frac{-q}{p}$must be a root of the second equation also.
• Since $\frac{-q}{p}$ is a root of the second equation, we can write:

$\begin{array}{ll}
{}&{d \left(\frac{-q}{p} \right)^2~+~2e\left(\frac{-q}{p} \right)~+~f}
&{~=~0}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{d q^2}{p^2}~-~\frac{2eq}{p}~+~f}
&{~=~0}& {}
&{\color {green} {\text{- - - - (a)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{d q^2}{p^2 q^2}~-~\frac{2eq}{p q^2}~+~\frac{f}{q^2}}
&{~=~\frac{d}{p^2}~-~\frac{2e}{p q}~+~\frac{f}{q^2}}& {~=~0}
&{\color {green} {\text{- - - - (b)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{p d}{p^2}~-~\frac{2ep}{p q}~+~\frac{f p}{q^2}}
&{~=~\frac{d}{p}~-~\frac{2e}{q}~+~\frac{f p}{q^2}}& {~=~0}
&{\color {green} {\text{- - - - (c)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{d}{p}~-~\frac{2e}{q}~+~\frac{f p}{pr}}
&{~=~\frac{d}{p}~-~\frac{2e}{q}~+~\frac{f}{r}}& {~=~0}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{2e}{q}}
&{~=~\frac{d}{p}~+~\frac{f}{r}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
(i) Line marked as (a):
We want to transform the statement in (a) to the staement in (1). For that, we divide by q²
(ii) Line marked as (b):
We want to further transform the statement in (b) to the statement in (1). For that, we multiply by p.
(iii) Line marked as (c):
From (2), we have: q² = pr
• Hence the statement in (1) is proved.

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The link below gives some more solved examples

Miscellaneous Exercise on chapter 9

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In the next chapter we will see straight lines.

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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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Chapter 9.3 - Geometric Progression

In the previous section, we completed a discussion on arithmetic progression. In this section, we will see geometric progression.

Some basics of geometric progression can be written in 6 steps:
1. In the case of arithmetic progression, we saw that:
We can subtract any term from the succeeding term. The difference will always be a constant.
2. In the case of geometric progression, we will see that:
We can divide any term by the preceding term. The ratio will always be a constant.
3. Let us see three examples:
Example 1:
• Consider the sequence: 2, 4, 8, 16, . . .
   ♦ Let us divide 4 by it’s preceding term 2. We will get: $\frac{4}{2}~=~2 $
   ♦ Let us divide 8 by it’s preceding term 4. We will get: $\frac{8}{4}~=~2 $
   ♦ Let us divide 16 by it’s preceding term 8. We will get: $\frac{16}{8}~=~2 $
• In this way, we can divide any term (except the first term) of this sequence by it’s preceding term. The result will be always 2
Example 2:
• Consider the sequence: $\frac{1}{9},\frac{-1}{27},\frac{1}{81},\frac{-1}{243},~.~.~.$
   ♦ Let us divide $\frac{-1}{27}$ by it’s preceding term $\frac{1}{9}$. We will get: $\frac{-1}{27} × \frac{9}{1}~=~\frac{-1}{3}$
   ♦ Let us divide $\frac{1}{81}$ by it’s preceding term $\frac{-1}{27}$. We will get: $\frac{1}{81} × \frac{27}{-1}~=~\frac{-1}{3}$
   ♦ Let us divide $\frac{-1}{243}$ by it’s preceding term $\frac{1}{81}$. We will get: $\frac{-1}{243} × \frac{81}{1}~=~\frac{-1}{3}$
• In this way, we can divide any term (except the first term) of this sequence by it’s preceding term. The result will be always $\frac{-1}{3}$
Example 3:
• Consider the sequence: 0.01, 0.0001, 0.000001, . . .
   ♦ Let us divide 0.0001 by it’s preceding term 0.01. We will get:
$\frac{0.0001}{0.01}~=~\frac{0.0001 × 100}{0.01 × 100}~=~\frac{0.01}{1}~=~ 0.01 $
   ♦ Let us divide 0.000001 by it’s preceding term 0.0001. We will get:
$\frac{0.000001}{0.0001}~=~\frac{0.000001 × 10000}{0.0001 × 10000}~=~\frac{0.01}{1}~=~ 0.01 $
• In this way, we can divide any term (except the first term) of this sequence by it’s preceding term. The result will be always 0.01
4. Note that, none of the terms in such sequences can be zero.
• The reason can be written in 5 steps:
(i) Suppose that one of the terms of the sequence is zero.
(ii) To find the constant ratio, we will want to divide the succeeding term of that zero by zero.
(iii) Division by zero will give a number that does not exist.
(iv) A number that does not exist cannot be the constant ratio.
(v) So we say that, none of the terms in such a sequence must be zero.
5. Now we can write the definition of geometric progression:
• A sequence a₁, a₂, a₃, . . . , a_n, . . . is called geometric progression, if each term is non-zero and $\frac{a_{k+1}}{a_k}~=~r~(\text{a constant})~\text{for}~k~\ge~1$
• A geometric progression is also called geometric sequence.
• Geometric progression is abbreviated as G.P
• The first term of a G.P is usually denoted by the letter a
• The constant ratio is denoted by the letter r. It is called the common ratio.
• The number of terms in a G.P is denoted by the letter n
• The last term of a G.P is denoted by the letter l
• The sum of n terms of a G.P is denoted by S_n
6. Based on the above information, a G.P will be in the form:
a, ar, ar², ar³, . . .

---

General term of a G.P

• This can be written in 6 steps:
1. We know that, first term of a G.P is a. That is., a₁ = a
• This can be written as: a₁ = a = ar⁰ = ar^1-1
2. We know that, second term of a G.P is ar. That is., a₂ = ar
• This can be written as: a₂ = ar = ar¹ = ar^2-1
3. Similarly, third term of a G.P is ar². That is., a₃ = ar²
• This can be written as: a₃ = ar² = ar^3-1
• so on . . .
4. Let us write the above results in order:
   ♦ a₁ = ar^1-1
   ♦ a₂ = ar^2-1
   ♦ a₃ = ar^3-1
   ♦ so on . . .
• We see a pattern. Based on that pattern, we can write:
   ♦ a_n = ar^n-1
• That means, n^th term of the G.P = ar^n-1
• This expression can be used as a formula to find the n^th term.
• To apply this formula, all we need to know are:
   ♦ The first term, a
   ♦ The common ratio, r
   ♦ The position of the term, n
5. Now we can write:
• A finite G.P will of the form: a, ar¹, ar², ar³, . . . , ar^n-1.
• An  infinite G.P will of the form: a, ar¹, ar², ar³, . . . , ar^n-1, . . .
6. We can write about geometric series also:
• A finite geometric series will of the form: a+ ar¹+ ar² + ar³ + . . . + ar^n-1.
• An infinite finite geometric series will of the form: a + ar¹ + ar² + ar³ + . . . + ar^n-1 + . . .

---

Sum of n terms of a G.P

Formula for sum, can be derived in 3 steps:
1. We have:
S_n = a+ ar¹+ ar² + ar³ + . . . + ar^n-1.
2. If r = 1, then the sum can be easily calculated:
S_n = a+ a(1)¹+ a(2)² + a(3)³ + . . . + a(4)^n-1.
⇒ S_n = a+ a + a + a + . . . + a. (n terms)
⇒ S_n = na
3. If r ≠ 1, then two steps are required:
(i) Multiply the expression in (1) by r. We get:
rS_n = ar + ar² + ar³ + . . . + arⁿ.
(ii) Subtract 3(i) from (1). We get:
$S_n - r S_n~=~a-a r^n$
⇒ $(1 - r) S_n~=~a(1- r^n)$
⇒ $S_n~=~\frac{a(1- r^n)}{1 - r}$
⇒ $S_n~=~\frac{a (r^n - 1)}{r-1}$
• This expression can be used as a formula to find the sum of n terms.

---

Solved example 9.9
Find the 10^th and n^th terms of the G.P. 5, 25,125, . . .
Solution:
1. We have: a_n = ar^n-1
2. In our present case:
a = 5, r = 25/5 = 5
3. So the 10^th term = 5 × 5^10-1 = 5 × 5⁹ = 5¹⁰
4. Similarly, the n^th term = 5 × 5^n-1 = 5ⁿ

Solved example 9.10
Which term of the G.P., 2,8,32, . . . up to n terms is 131072?
Solution:
1. We have: a_n = ar^n-1
2. In our present case:
a = 2, r = 8/2 = 4
3. Substituting the known values, we get:
131072 = 2 × 4^(n-1)
⇒ 65536 = 4^(n-1)
4. Factorization of 65536 is shown in fig.9.2 below:

Fig.9.2

• We can write:
$65536~=~4^5 × 8^2~=~4^5 × (4 × 2)^2 ~=~4^5 × 4^2 × 4 ~=~4^8~=~4^{n-1}$
• Thus we get: (n-1) = 8
⇒ n = 9
5. So 131072 is the 9^th term of the given G.P

Solved example 9.11
In a G.P, the 3^rd term is 24 and the 6^th term is 192. Find the 10^th term.
Solution:
1. Let a be the first term and r the common ratio.
• Then we can write: a_n = ar^n-1
2. Substituting the known values, we get:
(i) a₃ = a × r^3-1 = a × r² = 24
(ii) a₆ = a × r^6-1 = a × r⁵ = 192
3. Taking ratio, we get:
$\frac{a r^5}{a r^2}~=~\frac{192}{24}~=~8$
⇒ r³ = 8 = (2)³
⇒ r = 2
4. Substituting this value of r in 2(i), we get:
a × 2² = 24
⇒ a × 4 = 24
⇒ a = 6
5. Now we can write:
a₁₀ = a × r^10-1 = 6 × 2⁹ = 3072

Solved example 9.12
Find the sum of first n terms and the sum of first 5 terms of the geometric series:
$1+\frac{2}{3}+\frac{4}{9}+~.~.~.$
Solution:
1. In the G.P, a = 1 and r = $\frac{\frac{2}{3}}{1}~=~\frac{2}{3}$
2. We have: $S_n~=~\frac{a (r^n - 1)}{r-1}$
• Substituting the known values, we get:
$S_n~=~\frac{1 × \left[\left(\frac{2}{3} \right)^n - 1\right]}{\frac{2}{3}-1}$

⇒ $S_n~=~\frac{\frac{2^n - 3^n}{3^n}}{\frac{2-3}{3}}~=~\frac{\frac{3^n - 2^n}{3^n}}{\frac{3-2}{3}}~=~\frac{\frac{3^n - 2^n}{3^n}}{\frac{1}{3}}$

⇒ $S_n~=~\frac{3^n - 2^n}{3^n} × \frac{3}{1}~=~\frac{3^n - 2^n}{3^{n-1}}$

3. We can write the sum of first 5 terms using the above expression:
$S_5~=~\frac{3^5 - 2^5}{3^{5-1}}~=~\frac{3^5 - 2^5}{3^4}~=~\frac{211}{81}$

Solved example 9.13
How many terms of the G.P $3, \frac{3}{2}, \frac{3}{4},~.~.~.$ are needed to give the sum $\frac{3069}{512}$ ?
Solution:
1. In the G.P, a = 3 and r = $\frac{\frac{3}{2}}{3}~=~\frac{1}{2}$
2. We have: $S_n~=~\frac{a (r^n - 1)}{r-1}$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{\frac{3069}{512}}
&{}={}& {\frac{3 × \left[\left(\frac{1}{2} \right)^n - 1\right]}{\frac{1}{2}-1}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{\frac{3069}{512}}
&{}={}& {\frac{\frac{3 × (1^n - 2^n)}{2^n}}{\frac{1-2}{2}}}
&{}={}& {\frac{\frac{3 × (2^n - 1^n)}{2^n}}{\frac{2-1}{2}}}
&{}={}& {\frac{\frac{3 × (2^n - 1)}{2^n}}{\frac{1}{2}}}
&{}& {}&{}& {} &{} \\


{\Rightarrow}&{\frac{3069}{512}}
&{}={}& {\frac{3 × (2^n - 1)}{2^n} × \frac{2}{1}}
&{}={}& {\frac{6 × (2^n - 1)}{2^n}}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{\frac{3069}{512 × 6}}
&{}={}& {\frac{(2^n - 1)}{2^n}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{1023 × 2^n}
&{}={}& {1024 × 2^n~-1024}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{2^n}
&{}={}& {1024}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

3. Factorization of 1024 is shown in fig.9.3 below:

Fig.9.3

• We can write: $2^n~=~1024~=~4^2 × 8^2 ~=~2^4 × 2^6 ~=~2^{10}$
• Thus we get: n = 10
• That means, 10 terms are needed to obtain the given sum.

Solved example 9.14
The sum of first three terms of a G.P is $\frac{13}{12}$ and their product is -1. Find the common ratio and the terms.
Solution:
1. Let the first three terms be: $\frac{a}{r}, a~\text{and}~ar$
Then we can write:
(i) $\frac{a}{r}+a+ar~=~\frac{13}{12}$
(ii) $\frac{a}{r} × a × ar~=~a^3~=~-1$
2. From 1(ii), we get: a = -1
Substituting this value of a in 1(i), we get:
$\frac{-1}{r}+(-1)+(-1)r~=~\frac{13}{12}$
⇒ $\frac{-1}{r} - 1 - r~=~\frac{13}{12}$
3. The above expression can be simplified as follows:
$\begin{array}{ll}
{}&{\frac{-1}{r} - 1 - r}
&{}={}& {\frac{13}{12}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{\frac{-1-r-r^2}{r}}
&{}={}& {\frac{13}{12}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{-12-12r-12r^2}
&{}={}& {13r}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{12r^2 +25r + 12}
&{}={}& {0}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

4. Solving the above quadratic equation, we get:
$r~=~\frac{-3}{4},~\frac{-4}{3}$
5. Now we can write the G.P:
(i) $\text{When}~r~=~\frac{-3}{4},~\text{the G.P is:}~\frac{-1}{\frac{-3}{4}},-1, -1 × \frac{-3}{4}$
• That is., $\frac{4}{3}, -1, \frac{3}{4}$

(ii) $\text{When}~r~=~\frac{-4}{3},~\text{the G.P is:}~\frac{-1}{\frac{-4}{3}},-1, -1 × \frac{-4}{3}$
• That is., $\frac{3}{4}, -1, \frac{4}{3}$

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In the next section we will see two more solved examples. We will also see geometric mean.

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