Geometric Progression or GP

Geometric Progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant called the common ratio.

For Example, the sequence given below forms a GP with a common ratio of 2

1 2 4 8 16 . . . n
⇑ ⇑ ⇑ ⇑ ⇑ . . .
1^st 2^nd 3^rd 4^th 5^th . . . n^th Terms

Note: Geometric Progression is also known as Geometric Sequence.

Some Other Examples of GP

• All Same Numbers
  For example, 5, 5, 5, .... form a GP with a common ratio 1.
• Branches of a Perfect Tree
  For instance, if a tree branches into two new branches at each level, the number of branches can be represented as 1, 2, 4, 8, .. which forms a GP with a common ratio of 2.
• Return on Investment
  For example, if your invested money becomes 1.2x every year.
  If your initial investment is 10, then your money after every year will be 12, 14.4, 17.28, .... which forms a GP with common ratio 1.2

• 100, 50, 25, 12.5, 6.25 . . . is a GP with common ratio r = 0.5
• -3, 6, -12, 24, -48 . . . is a GP with common ratio -2
• 2, -2, 2, -2, ..... is a GP with common ratio r = -1

The below diagram shows a sequence in GP with a common ratio of 2.

Properties of Geometric Progression

Geometric Sequence has the following key properties:

• The square of a term in GP is a product of its adjacent terms. a²_k = a_k-1 × a_k+1
• In a finite GP, the product of the terms at the same distance from the beginning and the end is the same. It means, a₁ × a_n = a₂ × a_n-1 =...= a_k × a_n-k+1
• If we multiply or divide a non-zero quantity by each term of the GP, then the resulting sequence is also in GP with the same common difference. Please note that addition and subtraction do not keep a GP as GP. It is true for AP only.
• The reciprocal of all the terms in GP also forms a GP with a common ratio of 1/r
• 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.
• Geometric progressions typically mean either exponential growth or exponential decay, depending on whether the common ratio is 0 < r < 1 or r > 1. This makes it useful in modeling things like population growth, radioactive decay, and interest compounding.
• If we take the log of GP terms, we get an Arithmetic Progression or AP. For example, if the terms of the GP are a, ar, ar², ar³,… the logarithms of these terms will be log⁡a, log⁡a + log⁡r, log⁡a+2log⁡r, … which form an AP.
• If we select terms at regular intervals (say every kth term) from a GP, those terms also form a GP with a common ratio d^k. For example, if we select every third term from 1, 2, 4, 8, 16, 32, 64, ... we get 1, 8, 64, ... which is again a GP with common ratio as 8.

General Form of Geometric Progression

A geometric sequence is a series of numbers in which the ratio between two consecutive terms is constant. This ratio is known as the common ratio denoted by 'r', where r ≠ 0.

The n^th term of the Geometric series is denoted by a_n and the elements of the sequence are written as a₁, a₂, a_3, a₄, ..., a_n.

a₁ = a_,
a₂ = a*r
a₃ = a*r²
a₄ = a*r³
a_n = a*r^n-1

Conditions for the given sequence to be a geometric sequence:

For any sequence to be considered a GP, the ratio of any two successive terms must remain constant:

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

Geometric Progression Formula

The following table shows Key Formulas and Properties:

General Form of Geometric Progression

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 of a Geometric Series is given by:

r = successive term/preceding term = ar^{n-1} / ar^{n-2}

Nth Term of Geometric Progression

The terms of a GP are represented 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}

General term or nth term of a Geometric Sequence a, ar, ar², ar³, ar⁴ is given by : 

a_n = ar^n-1

where, 

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

Nth Term from the Last Term is given by:

a_n = l/r^n-1

where,
l is the last term

Geometric Progression Sum of N Terms

The geometric progression summation 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, the Sum of the First n Terms of a GP is given by:

S_n = a(1 - rⁿ)/(1 - r), if r < 1
S_n = a(rⁿ -1)/(r - 1), if r > 1

Sum of an Infinite Geometric Progression

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

Let us take a geometric sequence a, ar, ar², ..., which has infinite terms. S_∞ denotes the sum of the 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

Thus, the Sum of an Infinite Geometric Progression is given by,

S_∞= a/(1-r), where |r| < 1

Geometric Sequence 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.

For an example, let's look at the sequence: 1, 2, 4, 8, 16, 32

Recursive formula of Geometric Series is given by

term(n) = term(n - 1) × 2

To find any term, we 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

Read More:Recursive Formula

Types of GP

GP is further classified into two types, which are:

1. Finite Geometric Progression (Finite GP)
2. Infinite Geometric Progression (Infinite GP)

Finite Geometric Progression

A 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. 

An example 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 are:

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

Geometric Progression vs Arithmetic Progression

Here are the key differences between Geometric Progression and Arithmetic Progression :

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Solved Examples on Geometric Progression (GP)

Let's solve some example problems on Geometric sequence.

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

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.

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[31/1]
     = 1 × 31
     = 31

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

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

Practice Questions on GP

Try these practice questions on geometric progression.

1. What is the common ratio of the following sequence: 2, 6, 18, 54, ...? How can this be confirmed?
2. If the first term of a geometric progression is 3 and the common ratio is 0.5, what are the first five terms of the GP?
3. In a geometric progression where the first term is 5 and the common ratio is -2, find the 7th term of the sequence.
4. Calculate the sum of the first 6 terms of the geometric progression: 10, 20, 40, 80, 160, 320.
5. How would you derive the formula for the nth term of a geometric progression using the first term and the common ratio?
6. If the sum of the first n terms of a geometric progression is 120, the first term is 5, and the common ratio is 2, what is the value of n?
7. Determine the 9th term of a geometric progression that starts with 1 and has a common ratio of 3.
8. How can you find the sum of an infinite geometric progression with a first term of 2 and a common ratio of 0.5?

Suggested Quiz

10 Questions

If a geometric progression has a common ratio of 1/2, what type of sequence does it represent?

• A

  Linear growth

• B

  Exponential decay

• C

  Quadratic growth

• D

  Constant sequence

Explanation:

A geometric progression with a common ratio of r = 1/2​ represents exponential decay, as each term is a fixed fraction (r < 1) of the previous term, leading to a sequence that decreases exponentially.

 Which of the following statements about geometric progressions is true?

• A

  Adding a constant to all terms of a GP results in another GP.

• B

  Multiplying all terms of a GP by a non-zero constant does not alter the common ratio.

• C

  The product of terms equidistant from the beginning and end of a finite GP is always the same.

• D

  The reciprocal of all terms in a GP forms an arithmetic progression (AP).

Explanation:

In a finite GP, terms equidistant from the start and end always have the same product due to the symmetry of the common ratio's exponents.

If the 4th term of a GP is 16, and the common ratio is 2, what is the first term?

• A

  1

• B

  2

• C

  4

• D

  8

Explanation:

The formula for the nth term of a GP is: a_n = a ⋅ r^{n − 1}

Here:

• a₄ = 16 (4th term)
• r = 2 (common ratio)
• n = 4

Substitute into the formula:
16 = a ⋅ 2^{4 − 1}
16 = a ⋅ 2³
16 = a ⋅ 8
a = 4

What is the sum of an infinite geometric series when the first term is 'a' and the common ratio is 'r', where |r| < 1?

• A

  a / (1 + r)

• B

  a / (1 - r)

• C

  a * (1 - r)

• D

  a * r / (1 - r)

Explanation:

The sum of an infinite geometric series when the first term is a and the common ratio is r, where ∣r∣ < 1, is given by the formula:

S_∞ = a/1 − r

If the terms x, y, z form a GP, which of the following must be true?

• A

  y² = xz

• B

  z² = xy

• C

  x² = yz

• D

  xyz = y³

Explanation:

In a GP, the middle term is the geometric mean of the first and third terms, so y² = xz. This means the middle term squared equals the product of the outer terms.

A sequence is defined by a1 = 2 and an = 3an−1. What is the 5th term of the sequence?

• A

  18

• B

  545

• C

  162

• D

  486

Explanation:

To find the 5th term (a₅), calculate step by step:

• a₁ = 2
• a₂ = 3 ⋅ a₁ = 3 ⋅ 2 = 6
• a₃ = 3 ⋅ a₂ = 3 ⋅ 6 = 18
• a₄ = 3 ⋅ a₃=3 ⋅ 18 = 54
• a₅ = 3 ⋅ a₄ = 3 ⋅ 54 = 162

If a, b, c are the n-th, m-th, and k-th terms of a G.P., then what is the value of: (b/a)k ⋅ (cb)n ⋅ (ac)m.

• A

  1

• B

  1/2

• C

  2

• D

  0

Explanation:

Let the first term of the G.P. be A and the common ratio be r. The general term of a G.P. is: T_x = A ⋅ r^x−1

Expressing a, b, and c:

• a = A ⋅ rⁿ⁻¹
• b = A ⋅ r^m−1
• c = A ⋅ r^k−1

Substituting into the given expression:

[Tex]\frac{b}{a} = \frac{A⋅r^{ m−1 }}{ A⋅r^{ n−1}} ​ =r^{ m−n}[/Tex]

[Tex]\frac{c}{b} = \frac{A⋅r^{ k−1 }}{ A⋅r^{ m−1}} ​ =r^{ k−m}[/Tex]

[Tex]\frac{a}{c} = \frac{A⋅r^{ n−1 }}{ A⋅r^{ k−1}} ​ =r^{ n−k}[/Tex]

Now, compute the product:

[Tex](\frac{b}{a})^k\cdot (\frac{c}{b})^n\cdot(\frac{a}{c})^m= (r^{ m−n })^k\cdot (r^{ k−m })^n \cdot(r^{ n−k })^m [/Tex]

[Tex](r^{ m−n })^k\cdot (r^{ k−m })^n \cdot(r^{ n−k })^m [/Tex]

r^{km-kn-nm+nk+nm-mk}

km-kn-nm+nk+nm-mk = 0
thus:
r⁰ = 1

The geometric mean of two numbers is 8, and their arithmetic mean is 10. What are the numbers?

• A

  16, 4

• B

  20, 5

• C

  64, 8

• D

  100, 20

Explanation:

Let the two numbers be a and b.
Given:

1. Geometric Mean:√ab = 8  ⟹  ab = 64
2. Arithmetic Mean: (a + b)/2 = 10  ⟹  a + b = 20

Solve:
The numbers a and b satisfy the quadratic equation:

x² − (a + b)x + ab = 0

Substitute a + b = 20 and ab = 64: x² − 20x + 64 = 0

Factorise the quadratic equation: x² − 20x + 64 = (x − 16)(x − 4) = 0

Thus, the roots are: x = 16 and x = 4

If three geometric means are inserted between 3 and 243, find the second geometric mean.

• A

  27

• B

  23

• C

  24

• D

  34

Explanation:

The GP will be 3, g₁, g₂, g₃, 243, where: a = 3, ar = g₁, ar² = g₂, ar³ = g₃, ar⁴ = 243

From the last term:
3 ⋅ r⁴ = 243
r⁴ = 243/3 = 81  ⟹  r = 3

Now, the second geometric mean is: g₂ = ar² = 3 ⋅ 3² = 3 ⋅ 9 = 27

Find the sum of the series 7 + 77 + 777 + … up to n terms.

• A

  S_n = 7/81(10ⁿ⁺¹ - 10 - 9n)

• B

  S_n = 7/81(10^n-1 - 10 - 9n)

• C

  S_n = 7/81(10ⁿ⁺¹ - 10 - 7n)

• D

  S_n = 8/81(10ⁿ⁺¹ - 10 - 9n)

Explanation:

Each term can be represented as repeated digits of 7. The k-th term of the series is:

T_k = 7 ⋅ (10^k−1 + 10^k−2 + ⋯ + 10⁰)
T_k ​ = 7⋅10^k−1​/9
Total Sum(S_n): S_n​ = T₁ ​+ T₂​ + ⋯ + T_n​
Substitute T_k:
Sn​ = 7/9​ ⋅ [(10¹ + 10² + ⋯ + 10ⁿ) − n]

The sum of powers of 10 is another geometric progression:
10¹ + 10² + ⋯ + 10ⁿ = 10(10ⁿ − 1)/9

​Substitute this into the equation:
S_n ​= 7/81​ ⋅ [10(10ⁿ − 1) − 9n]

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