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, . . .

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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 + . . .

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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.

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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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