# 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:

a

_{1}, a_{2}, a_{3,}a_{4}, …, a_{n}

Given sequence is a geometric sequence if:

a

_{2}/a_{1}= a_{3}/a_{2}= … = a_{n}/a_{n-1 }= r (common ratio)

The given sequence can also be written as:

a, ar, ar

^{2}, ar^{3}, … , 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, ar^{2}, ar^{3},……ar^{n-1}, ar^{n}.

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, ar^{2}, ar^{3},……ar^{n-1}, ar^{n}……, 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[(r^{n}– 1)/(r – 1)] if r ≠ 1and r > 1 - S
_{n}= a[(1 – r^{n})/(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, b^{2}= 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, ar^{2}, ar^{3}, ar^{4}………. 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 a_{1}, a_{2}, a_{3}, a_{4}, …, a_{n}, expressing all these terms according to the first term a_{1}, we get

a

_{1 }= a

a_{2 }= a_{1}r

a_{3 }= a_{2}r = (a_{1}r)r = a_{1}r^{2}

a_{4 }= a_{3}r = (a_{1}r^{2})r = a_{1}r^{3}

…

a_{m }= a_{1}r^{m−1}

…

Similarly,a_{n }= a_{1}r^{n – 1}

a_{n }= ar^{n – 1}

where,a

_{1}= first term,

a_{2}= 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,

lis 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, if r < 1_{n}= a(1 – r^{n})/(1 – r)

S, if r > 1_{n}= a(r^{n}-1)/(r – 1)

### Derivation of the Formula

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

S = a_{1 }+ a_{2 }+ a_{3 }+ … + a_{n}

S = a_{1 }+ a_{1}r + a_{1}r^{2 }+ a_{1}r^{3 }+ … + a_{1}r^{n−1} ….equation (1)

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

S × r= a_{1}r + a_{1}r^{2 }+^{ }a_{1}r^{3 }+ a_{1}r^{4 }+ … + a_{1}r^{n} ….equation (2)

Subtract Equation (2) from Equation (1)

S – Sr = a_{1} – a_{1}r^{n}

(1 – r)S = a_{1}(1 – r^{n})

S_{n} = a_{1}(1 – r^{n})/(1 – r), if r<1

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

Sr – S = a_{1}r^{n }–^{ }a_{1}

(r – 1)S = a_{1}(r^{n}-1)

Hence,

S_{n}= a_{1}(r^{n}-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 – r^{n})/(1 – r)

S = (a – ar^{n})/(1 – r)

S = a/(1 – r) – ar^{n}/(1 – r)

For n -> ∞, the quantity (ar^{n}) / (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, ar^{2}, … which has infinite terms. S_{∞} denotes the sum of infinite terms of that sequence, then

S_{∞} = a + ar + ar^{2 }+ ar^{3}+ … + ar^{n }+..(1)

Multiply both sides by r,

rS_{∞} = ar + ar^{2 }+ ar^{3}+ … … (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

- a
^{2}_{k}= a_{k-1 }× a_{k+1} - a
_{1}× a_{n}= a_{2}× 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 y
^{2}= 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 a

_{1}term(n) = term(n – 1) × r

= term(n -1) × 1.5 for n>=2

a

_{1 }= 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, ar^{2}, ar^{3}, ar^{4}

a = 4

ar = 4 × 5 = 20

ar^{2}= 4 × 25 = 100

ar^{3}= 4 × 125 = 500

ar^{4}= 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[(r^{n}– 1)/(r – 1)]

S_{5}= 1[(2^{5}– 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,

a_{9}= 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(r^{n}-1)/(r – 1) = 1(2^{6}-1)/(2-1) = 63

**Example: Given a geometric sequence with a _{1 }= 3 and a_{4 }= 24, find a_{5}**

**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 a

_{1 }and r. Substitute 24 for a_{4. }Solve for the common ratio.a

_{n}= a_{1}× r^{n-1}a

_{4}= 3r^{3}24 = 3r

^{3}8 = r

^{3}r = 2

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

a

_{5 }=_{ }a_{1}× 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, ar

^{2}, ar^{3}, ar^{4},…,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, ar

^{2}, ar^{3},……ar^{n-1}, then the formula to find the sum of GP is:Sn = a + ar + ar

^{2}+ ar^{3}+…+ ar^{n-1}

Sn = a[(r^{n}– 1)/(r – 1)]where r ≠ 1 and r > 1

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