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Geometric Sequence

What is a Sequence?

In order to study geometric sequence, it is important for us to understand what we mean by a sequence? 

A sequence is a function whose domain is the set N of natural numbers.

It is compulsory to denote a sequence by a letter ‘a’ and the image a(n) of n ∈ under a by an. since the domain for every sequence is the set N of natural numbers, therefore, a sequence is represented by its range. The images of 1, 2, 3 ….., n…. under a sequence ‘a’ are denoted by a1, a2, a3, ……… an respectively. a1, a2, a3, ……… an are known as first term, second term…… nth term respectively of the sequence. If an is the nth term of a sequence, ‘a’ then we write a = <an>.

What is a Real Sequence?

A sequence whose range is a subset of R is called a real sequence. In other words, a real sequence is a function with domain N and the range a subset of the set R of real numbers

Representation of a Sequence

There are several ways of representing a real sequence. One way to represent a real sequence is to list its first few terms till the rule for writing down other terms becomes clear, for example, 1, 3, 5 ….. is a sequence whose nth term is ( 2n – 1 ).

Another way of representing a real sequence is to give a rule of writing the nth term of a sequence. For example, the sequence, 1, 3, 5, 7 ….. can be written as an = 2n – 1.

Sometime we represent a real sequence by using a recursive relation. For example, the Fibonacci sequence is given by 

a1 = 1, a2 = 1 and an + 1 = an + an – 1, n ≥ 2

The terms of this sequence  1, 1, 2, 3, 5, 8, ……..

Let us now understand what we mean by a geometric sequence.

What is a Geometric Sequence?

A sequence of non-zero numbers is called a geometric sequence, also known as geometric progression (G. P ) if the ratio of a term and the term preceding it is always a constant quantity.

The constant ratio is called the common ratio of the geometric sequence.

In other words, a sequence, a1, a2, a3, ……… an is called a geometric progression if $\frac{a_n+ 1}{a_n}$ = constant for all n belonging to N.

For example,

The sequence, 4, 12, 36, 108,….. is a geometric sequence because $\frac{12}{4} = \frac{36}{12} = \frac{108}{36}$ = ….. = 3 which is constant. Similarly,

The sequence, $\frac{1}{3}, \frac{-1}{2}, \frac{3}{4}, \frac{-9}{8}$, …….. is a geometric sequence with the first term $\frac{1}{3}$ and common ratio equal to $\frac{-1}{2} ÷ \frac{1}{3}  = \frac{-3}{2}$

So, now we can define a geometric series as – 

If a1, a2, a3, ……… an is a geometric sequence then the expression a1 + a2 +a3 +…… an + …… is called a geometric series.

Let us now understand what we mean by finite and infinite geometric sequence.

Finite Geometric Sequence

Finite geometric progression is the geometric series that contains a finite number of terms. In other words, it is the sequence where the last term is defined. For example, the sequence,  3, 6, 12, 24, 3072 is a finite geometric sequence having the first term 3 and last term 3072, with a common ratio 2.

Infinite Geometric Sequence

Infinite geometric progression is the geometric sequence that contains an infinite number of terms. In other words, it is the sequence where the last term is not defined.  For example, 2 ,6, 15, 54, ….. is an infinite geometric sequence, having the first term 2, common ratio 3 and no last term as the sequence is endless. 

General Term of a Geometric Sequence

Let a1, a2, a3, ……… an be the given geometric sequence with first term a and common ratio r. Then, we have

a1 = a ⇒ a1 = ar 1 – 1

Since, a1, a2, a3, ……… an is a geometric sequence with common ratio r, this means that 

$\frac{a_2}{a_1}$ = r

This implies that,

a2 = r a1

which further means that 

a2 = a r

Hence, we have,

a2 = a r 2 – 1

Also,

$\frac{a_3}{a_2}$ = r

This implies that,

a3 = r a2

which further means that 

a3 = ( a r ) r 

⇒ a3 = a r 2

which further implies that

 a3 = a r 3 – 1

We will now find the value of $\frac{a_4}{a_3}$

We know that

$\frac{a_4}{a_3}$ = r

Therefore, this means that

a4 = r a3

which further means that 

a4 = a r 3

Finally,

a4 = a r 4 – 1

If we continue in this manner, we can come up to the conclusion that an = a r n – 1

So, from the above calculations, it can be concluded that,

If a is the first term and r is the constant ratio of a geometric sequence, then the geometric sequence can be written as a, ar, ar2, ar3, ……. ar n – 1  or a, ar, ar2, ar3, ar4, …….. ar n – 1, ….. according to as it is finite or infinite.

The nth term of a Geometric Sequence

Now, let us define the nth term of a geometric sequence.

Since the geometric sequence consists of m terms.

Therefore, the nth term of the sequence from the end will be equal to the ( m – n + 1 ) th term from the beginning of the sequence which will further be equal to ar m – n

Hence, we can say that,

The nth term from the end of a finite geometric sequence, consisting of m terms is equal to ar m – n, where a is the first term and r is the common ratio of the geometric sequence.

Now, what would be the nth term of a geometric sequence with the last term l and common ratio r  . Let us find out.

Clearly, when we look at the terms of a geometric sequence from the last term and move towards the beginning we find that the progression is a geometric sequence with common ratio 1r. Therefore, the nth term from the end is given by an = l ($\frac{1}{r}$)  n – 1

Hence, we can say that,

The nth term of a geometric sequence having the  last term l and common ratio r is given by 

an = l ($\frac{1}{r}$)  n – 1.

Examining Geometric Series under Different Conditions

Let us now understand how to solve problems of the geometric sequence under different conditions.

Finding the indicated Term of a Geometric Sequence when its first term and the common ratio are given

Example Find the 4th term and the general term of the sequence, 3, 6, 12, 24, ………. , 3072.

Solution We have been given the geometric sequence, 3, 6, 12, 24, ………. , 3072.

We are required to find the 4th term from the end of this sequence.

Let us first identify the nth term of this sequence.

Clearly, the common ratio of the given sequence is 2.

We also know that its nth term will be 

an = l ($\frac{1}{r}$)  n – 1

Therefore, by substituting n by 4 in the above formula, we will get,

4th term of the sequence from the end =  l ($\frac{1}{r}$)  4 – 1

= 3072 ($\frac{1}{2}$)  4 – 1 

= $\frac{3072}{8}$

= 384

Hence, the 4th term and the general term of the sequence, 3, 6, 12, 24, ………. , 3072 will be 384.

Finding the Position of a Given Term in a Given Geometric Sequence

Example Which term of the geometric sequence is  5, 10, 20, 40, ……. Is 5120 ?

Solution We have been given the geometric sequence 5, 10, 20, 40, …….. We need to find the position of the term 5120 in this series.
To find the position of the given term, let us first find the first term and the common ratio of the sequence.

From the above sequence, we can see that the first term of the sequence is 5 and the ratio of the given sequence is 2.

Hence, a = 5 and r = 2.

Now, let the nth term of the sequence be 5120.

Then, we have,

an = 5120

This means that,

a r n – 1 = 5120

⇒  5 ( 2 n – 1 ) = 5120

⇒ 2 n – 1  = 1024

⇒ 2 n – 1  = 2 10

⇒ n – 1 = 10

n = 11

Thus 5120 will be the 11 th term of the sequence 5, 10, 20, 40, …….

Finding the Sum of a Given Number of Terms of a Given geometric Sequence 

Let Sn denote the sum of n terms of the geometric sequence with first term ‘a’ and common ratio ‘ ‘r‘.  Then, we have,

Sn = a + ar + ar2, ar3, ……. ar n – 1 ……………………………. ( 1 )

Multiplying both sides by r we get,

 r Sn  = + ar + ar2, ar3, ……. ar n – 1 + an …………………………….. ( 2 )

On subtracting ( 2 ) from (  1 ) we get,

Sn – r Sn   = a – a r n

⇒ Sn  ( 1 – r ) =  a ( 1 – r n )

⇒ Sn =  $\frac{a(1 – r^n)}{1-r}$, provided r ≠ 1

Or,

Sn =  a [$\frac{r^n- 1}{r-1}$] 

Therefore, we have, Sn =  $\frac{a(1 – r^n)}{1-r}$ , provided r ≠ 1 or Sn =  a [$\frac{r^n- 1}{r-1}$]

Hence, we can say that,

The sum of n terms of a geometric sequence with the first term ‘ a ‘ and common ratio ‘ r  ‘  is given by  

Sn $\frac{a(1 – r^n)}{1-r}$, provided r ≠ 1 or Sn =  a [$\frac{r^n- 1}{r-1}$] 

Let us understand this through an example.

Example Find the sum of 7 terms of the geometric sequence, 3, 6, 12 , ……..

Solution We have been given the geometric series 3, 6, 12 , ……..

The first term of the sequence is 3 and the common ratio of the sequence is 2 and the number of terms in the sequence whose sum is to be obtained is 7

So, we have,

a = 3, r = 2 and n = 7 

Now, we have learnt that for a geometric sequence with the first term ‘ a ‘ and common ratio ‘ r  ‘  , the sum of n terms is given by

Sn =  a [$\frac{r^n- 1}{r-1}$]

Substituting the given values in the above equation, we have,

Sn =  3 [$\frac{2^7- 1}{2-1}$] = 3 ( 128 – 1 ) = 381

Hence, the sum of 7 terms of the geometric sequence, 3, 6, 12 , …….. is 381.

Applications of Geometric Sequence in Real Life 

The following are the general applications of geometric sequence in real life – 

  1. The geometric sequences have played an important role in the early development of calculus, and continue to play a central part in the study of the convergence of series. 
  2. Geometric sequences are used throughout mathematics. They have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance.
  3. Repeating Decimal – A repeating decimal can be thought of as a geometric series whose common ratio is a power of $\frac{1}{10}$. 
  4. Archimedes’ Quadrature of the Parabola Archimedes used the sum of a geometric sequence to compute the area enclosed by a parabola and a straight line. His method was to dissect the area into an infinite number of triangles. 
  5. Fractal Geometry – The Koch snowflake is a fractal shape with an interior comprised of an infinite amount of triangles. In the study of fractals, geometric sequences often come up as the perimeter, area, or volume of a self-similar figure. In the case of the Koch snowflake, its area can be described with a geometric series.

Solved Examples

Example 1 The 4 th and the 8th term of a geometric sequence are 8 and 128 respectively. Find the geometric sequence.

Solution We have been given that the 4 th and the 8th term of a geometric sequence are 8 and 128 respectively. We need to find the respective geometric sequence. 

Let a be the first term and r be the common ratio of the geometric sequence. 

Since the 4 th term of the geometric sequence is 8, therefore, we have,

ar3 = 8 ……….. ( 1 )
Also, the 8th term of the geometric sequence is 128, therefore we have,

ar7 = 128 ………… ( 2 )

We shall now divide equation 2 by equation 1. We will now have, 

r5 = 16 which implies that r = 2

On substituting the value of r in the equation 1, we will get,

a23 = 8

⇒ a = 1

Hence, the geometric sequence is defined as 1, 2, 4, 8 , 16, ……. 

Example 2 Find the 15th term of a G.P whose 8th term is 192 and the common ratio is ‘2’

Solution Let first term of G.P is ‘a’ and common ratio r = 2

8th term of G.P is 192 So

⇒ 192 = a x (2)7

⇒ a = 192 / (2)7

Now 15th is

a15  = [ 192 / (2)7 ] (2) 14

a15  =  192 x 2 7   = 3 x 2 x 2 7   = 3 x 2 13

Hence, the 15th term of a G.P whose 8th term is 192 and the common ratio is ‘2’ is 3 x 2 13

Remember 

  1. A sequence is a function whose domain is the set N of natural numbers.
  2. A sequence whose range is a subset of R is called a real sequence.
  3. If a1, a2, a3, ……… an is a geometric sequence then the expression a1 + a2 +a3 +…… an + …… is called a geometric series.
  4. If a is the first term and r is the constant ratio of a geometric sequence, then the geometric sequence can be written as a, ar, ar2, ar3, ……. ar n – 1  or a, ar, ar2, ar3, ar4, …….. ar n – 1, ….. according to as it is finite or infinite.
  5. The nth term from the end of a finite geometric sequence, consisting of m terms is equal to ar m – n, where a is the first term and r is the common ratio of the geometric sequence.
  6. The nth term of a geometric sequence with last term l and common ratio r  is given by an = l ( $\frac{1}{r}$ )  n – 1
  7. The sum of n terms of a geometric sequence with the first term ‘ a ‘ and common ratio ‘ r  ‘  is given by  Sn =  $\frac{a(1 – r^n)}{1-r}$, provided r ≠ 1 or Sn =  a [$\frac{r^n- 1}{r-1}$]
  8. Finite geometric progression is the geometric series that contains a finite number of terms.
  9. Infinite geometric progression is the geometric sequence that contains an infinite number of terms.
  10. Geometric sequence has played an important role in the early development of calculus, and continues to play a central part of the study of the convergence of series. 
  11. Geometric sequences are used throughout mathematics. They have important applications in physics, engineering, biology, economics, computer science, queuing theory, and finance.
  12. Applications of geometric sequence are found in many mathematical concepts such as repeating decimals, Archimedes’ Quadrature of the Parabola and Fractal geometry.

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