Geometric Progression

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, a_3, a₄, …, a_n

Given sequence is a geometric sequence if:

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

The given sequence can also be written as:

a, ar, ar², ar³, … , ar^n-1  

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

The common ratio is given by:

r = successive term/preceding term = ar^n-1 / ar^n-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,……ar^n-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,……ar^n-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.

• n^th term of a GP is T_n = ar^n-1
• Common ratio = r = T_n/ T_n-1

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

• S_n = a[(rⁿ – 1)/(r – 1)] if r ≠ 1and r > 1
• S_n = a[(1 – rⁿ)/(1 – r)] if r ≠ 1 and r < 1
• n^th 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, k^th term from the end of the GP will be T_k = ar^n-k.

N^th 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 n^th term is denoted by a_n. Thus, to find the nth term of a Geometric Sequence will be : 

a_n = ar^n-1

Derivation of the Formula

Given each term of GP as a₁, a₂, a₃, a₄, …, a_n, 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
…
a_m = a₁r^m−1
…
Similarly,
a_n = a₁r^{n – 1}

a_n = ar^{n – 1}

where, 

a₁ = first term, 
a₂ = second term
a_n = last term (or the nth term)
a_m = any term before the last term

n^th term from the last term is given by:

a_n = l/r^n-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:

S_n = a(1 – rⁿ)/(1 – r), if r < 1

S_n = 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 + … + a_n

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)

S_n = 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, 

S_n = 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

• a²_k = a_k-1 × a_k+1
• a₁ × a_n = a₂ × a_n-1 =…= a_k × a_n-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;
S_n = 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:

a_n = ar^n-1

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

Example 4: Find the 6^th 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 = ar^n-1 = 1.2^6-1 = 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 a_4. Solve for the common ratio.

a_n = a₁ × r^n-1

a₄ = 3r³

24 = 3r³

8 = r³

r = 2

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

a₅ = a₁ × r^n-1

= 3 × 2^5-1

= 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,…,ar^n-1

a = First term
r = common ratio
ar^n-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,……ar^n-1, then the formula to find the sum of GP is:

Sn = a + ar + ar² + ar³ +…+ ar^n-1

Sn = a[(rⁿ – 1)/(r – 1)]

where r ≠ 1 and r > 1