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Geometric Progression

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  • Last Updated : 30 Sep, 2022
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Geometric Progression (GP) is a specific type of progression or sequence, where each next term in the progression is produced by multiplying the previous term by a fixed number, and the fixed number is called the Common Ratio. Similar to arithmetic progression, geometric progression also carries a specific pattern that is useful in dealing with GP questions. Common ratio and the first term of a GP is always a non-zero number. 
Example: 3,6,12,24,48, and so on is a GP with first term 3 and common difference 2.

What is a Geometric Progression?

A geometric sequence is one in which the ratio between two consecutive terms is constant. This ratio is known as the common ratio denoted by ‘r’, where r ≠ 0. Let the elements of the sequence be denoted by:

 a1, a2, a3, a4, …, an

Given sequence is a geometric sequence if:

a2/a1 = a3/a2 =  … = an/an-1 = r (common ratio)

The given sequence can also be written as:

a, ar, ar2, ar3, … , arn-1  

Here, r is the common ratio and a is the scale factor

The common ratio is given by:

r = successive term/preceding term = arn-1 / arn-2

Types of Geometric Progression

Geometric progression is further classified on the basis of whether they are ending or continuing infinitely. So, a GP is further classified into two parts which are:

  • Finite Geometric Progression (Finite GP)
  • Infinite Geometric Progression (Infinite GP)

The two types of GP are further explained below in this article

Finite Geometric Progression

Finite G.P. is a sequence that contains finite terms in a sequence and can be written as a, ar, ar2, ar3,……arn-1, arn

Examples of Finite GP is 1, 2, 4, 8, 16,……512

Infinite Geometric Progression

Infinite G.P. is a sequence that contains infinite terms in a sequence and can be written as a, ar, ar2, ar3,……arn-1, arn……, i.e. it is a sequence that never ends.

Examples of Infinite GP is 

  • 1, 2, 4, 8, 16,……..
  • 1, 1/2, 1/4, 1/8, 1/16,………

Geometric Progression Formulas

The list of formulas related to GP is given below which will help in solving different types of problems. The general form of terms of a GP is a, ar, ar2, ar3, and so on. Here, a is the first term and r is the common ratio.

  • nth term of a GP is Tn = arn-1
  • Common ratio = r = Tn/ Tn-1

The formula to calculate the sum of the first n terms of a GP is given by:

  • Sn = a[(rn – 1)/(r – 1)] if r ≠ 1and r > 1
  • Sn = a[(1 – rn)/(1 – r)] if r ≠ 1 and r < 1
  • nth term from the end of the GP with the last term l and common ratio r = l/ [r(n – 1)]
  • Sum of infinite, i.e. the sum of a GP with infinite terms is S= a/(1 – r) such that 0 < r < 1.

For three quantities in GP, the middle quantity is called the Geometric Mean of the other two terms. If a, b and c are three quantities in GP, then and b is the geometric mean of a and c then,

Now,  b2 = ac or b =√(ac)

If a is the first term and r is the common ratio respectively of a finite GP with n terms. Then, kth term from the end of the GP will be Tk = arn-k.

Nth Term of a Geometric Progression

To find the nth term of a Geometric Sequence, if the given series is in the form of a, ar, ar2, ar3, ar4………. then

The nth term is denoted by an. Thus, to find the nth term of a Geometric Sequence will be : 

an = arn-1

Derivation of the Formula

Given each term of GP as a1, a2, a3, a4, …, an, expressing all these terms according to the first term a1, we get

a1 = a
a2 = a1r
a3 = a2r = (a1r)r = a1r2
a4 = a3r = (a1r2)r = a1r3

am = a1rm−1

Similarly,
an = a1rn – 1

an = arn – 1

where, 

a1 = first term, 
a2 = second term
an = last term (or the nth term)
am = any term before the last term

nth term from the last term is given by:

an = l/rn-1

where,
l is the last term

Sum of the First n Terms of a Geometric Progression

Sum of the First n Terms of a Geometric Sequence is given by:

Sn = a(1 – rn)/(1 – r), if r < 1

Sn = a(rn -1)/(r – 1), if r > 1

Derivation of the Formula

The sum in geometric progression (known as geometric series) is given by

S = a1 + a2 + a3 + … + an

S = a1 + a1r + a1r2 + a1r3 + … + a1rn−1     ….equation (1)

Multiply both sides of Equation (1) by r (common ratio), and we get

S × r= a1r + a1r2 + a1r3 + a1r4 + … + a1rn     ….equation (2)

Subtract Equation (2) from Equation (1)

S – Sr = a1 – a1rn

(1 – r)S = a1(1 – rn)

Sn = a1(1 – rn)/(1 – r), if r<1

Now, Subtracting Equation (1) from Equation (2) will give

Sr – S = a1rn a1

(r – 1)S = a1(rn-1)

Hence, 

Sn = a1(rn -1)/(r – 1), if r > 1

Geometric Progression Sum to Infinite Terms

The number of terms in infinite geometric progression will approach infinity (n = ∞). The sum of infinite geometric progression can only be defined at the range of |r| < 1.

S = a(1 – rn)/(1 – r)

S = (a – arn)/(1 – r)

S = a/(1 – r) – arn/(1 – r)

For n -> ∞, the quantity (arn) / (1 – r) → 0 for |r| < 1, 

Thus,

S= a/(1-r), where |r| < 1

Derivation of Sum of Infinite Geometric Progression

Take a geometric sequence a, ar, ar2, … which has infinite terms. S denotes the sum of infinite terms of that sequence, then

S = a + ar + ar2 + ar3+ … + arn +..(1)

Multiply both sides by r,

rS = ar + ar2 + ar3+ … … (2)

subtracting eq (2) from eq (1),

S – rS = a

S (1 – r) = a

S = a / (1 – r)

Thus by the above formula sum of infinite terms of an infinite GP is found, 

Note: This formula only works if |r| < 1

Properties of Geometric Progression

  • a2k = ak-1 × ak+1
  • a1 × an = a2 × an-1 =…= ak × an-k+1
  • If we multiply or divide a non-zero quantity to each term of the GP, then the resulting sequence is also in GP with the same common difference.
  • Reciprocal of all the terms in GP also forms a GP.
  • If all the terms in a GP are raised to the same power, then the new series is also in GP.
  • If y2 = xz, then the three non-zero terms x, y, and z are in GP.

Recursive Formula

A recursive formula defines the terms of a sequence in relation to the previous value. As opposed to an explicit formula, which defines it in relation to the term number.

As a simple example, let’s look at the sequence: 1, 2, 4, 8, 16, 32

The pattern is to multiply 2 repeatedly. So the recursive formula is

term(n) = term(n – 1) × 2

Notice, in order to find any term you must know the previous one. Each term is the product of the common ratio and the previous term.

term(n) = term(n – 1) × r

Example: Write a recursive formula for the following geometric sequence: 8, 12, 18, 27, … 

Solution: 

The first term is given as 6. The common ratio can be found by dividing the second term by the first term.

r = 12/8 = 1.5

Substitute the common ratio into the recursive formula for geometric sequences and define  a1

term(n) = term(n – 1) × r 

= term(n -1) × 1.5 for n>=2

a1 = 6

Solved Example on Geometric Progression

Example 1: Suppose the first term of a GP is 4 and the common ratio is 5, then the first five terms of GP are?

Solution: 

First term, a = 4
Common ratio, r = 5
Now, the first five term of GP is
a, ar, ar2, ar3, ar4
a = 4
ar = 4 × 5 = 20
ar2 = 4 × 25 = 100
ar3 = 4 × 125 = 500
ar4 = 4 × 625 = 2500
Thus, the first five terms of GP with first term 4 and common ratio 5 are:
4, 20, 100, 500, and 2500

Example 2: Find the sum of GP: 1, 2, 4, 8, and 16.

Solution: 

Given GP is 1, 2, 4, 8 and 16
First term, a = 1
Common ratio, r = 2/1 = 2 > 1
Number of terms, n = 5
Sum of GP is given by;
Sn = a[(rn – 1)/(r – 1)]
S5 = 1[(25 – 1)/(2 – 1)]
     = 1[(32 – 1)/1]
     = 1[31/2]
     = 1 × 15.5
     = 15.5

Example 3: If 3, 9, 27,…., is the GP, then find its 9th term.

Solution: 

nth term of GP is given by:

an = arn-1

given, GP 3, 9, 27,….
Here, a = 3 and r = 9/3 = 3
Therefore,
a9 = 3 x 39 – 1
    = 3 × 6561
    = 19683

Example 4: Find the 6th term and sum of 6 terms of the Sequence: 1, 2, 4, 8, 16, 32 

Solution: 

Given Sequence, 1, 2, 4, 8, 16, 32
Common ratio r = 2/1 = 2
first term = 1
6th term in the sequence = arn-1 = 1.26-1 = 63
Sum of first 6 terms = a(rn -1)/(r – 1) = 1(26-1)/(2-1) = 63

Example: Given a geometric sequence with a1 = 3 and a4 = 24, find a5

Solution:

The sequence can be written in terms of the initial term and the common ratio r.

Write the fourth term of sequence in terms of  a1 and r.  Substitute 24 for a4. Solve for the common ratio.

an = a1 × rn-1

a4 = 3r3

24 = 3r3

8 = r3

r = 2

Find the second term by multiplying the first term by the common ratio.

a5 = a1 × rn-1

= 3 × 25-1

= 3 × 16 = 48

FAQs on Geometric Progression

Question 1: What is a Geometric Progression?

Answer:

Geometric Progression (GP) is a specific type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed constant, which is termed a common ratio(r).

For example, 1, 3, 9, 27, 81, …….

Question 2: What do you mean by the common ratio in GP?

Answer:

Common multiple between each successive term in a GP is termed the common ratio. It is a constant that is multiplied by each term to get the next term in the GP. If a is the first term and ar is the next term, then the common ratio is equal to:
ar/a = r

Question 3: Write the general form of GP.

Answer:

General form of a Geometric Progression (GP) is a, ar, ar2, ar3, ar4,…,arn-1

a = First term
r = common ratio
arn-1 = nth term

Question 4: What is the formula for the sum of a GP?

Answer:

Suppose the given GP is a, ar, ar2, ar3,……arn-1, then the formula to find the sum of GP is:

Sn = a + ar + ar2 + ar3 +…+ arn-1

Sn = a[(rn – 1)/(r – 1)]

where r ≠ 1 and r > 1


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