# Geometric Sequence and Series

• Last Updated : 21 Dec, 2020

A sequence is defined as an arrangement of numbers in a particular order, i.e., an ordered list of numbers. For example: 1, 3, 5, 7, … etc.

There are 2 types of sequences:

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Arithmetic sequence: An arithmetic sequence is the one in which the difference between two consecutive terms is constant. This difference is known as common difference.

Geometric sequence: In contrast, the geometric sequence is the one in which the ratio between two consecutive terms is constant. This ratio is known as common ratio.

## Series

A series is defined as the sum of the elements of a sequence. For example: 1 + 4 + 7 + 10 + … etc.

A series is of two types:

Finite Series: A finite series is one in which the number of elements in the series is known.

Infinite Series: When the number of elements in the series is not known, i.e. series with an infinite number of elements is known as infinite series.

## Geometric Sequence

A geometric sequence is the 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 are denoted by:

a1, a2, a3, a4, …, an

Given sequence is a geometric sequence if:

a1/a2 = a2/a3 = a3/a4 = … = an-1/an = 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

Common ratio is given by:

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

## What is the Nth Term of a Geometric Sequence?

To find the nth term of a Geometric Sequence, we know that the series is in the form of a, ar, ar2, ar3, ar4……….

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 = a1

a2 = a1r

a3 = a2r = (a1r)r = a1r2

a4 = a3r = (a1r2)r = a1r3

am = a1rm−1

an = a1rn – 1

where,

a1 = the first term, a2 = the second term, and so on

an: the last term (or the nth term) and

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

## What is the sum of the First n Terms of a Geometric Sequence?

Sum of 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), 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

### Sum of 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

Problem 1: Find Common Ratio and Scale Factor of the Sequence: 4, 12, 36, 108, 324, …

Solution:

Sequence provided is 4, 12, 36, 108, 324, …

Common ratio = 12/4 = 3

Scale factor = 4

Problem 2: Find Common Ratio and Scale Factor of the Sequence: 5, -5, 5, -5, 5, -5, …

Solution:

Given Sequence, 5, -5, 5, -5, 5, -5, …

Common ratio = -5/5 = -1

Scale factor = 5

Problem 3: Find nth Term and Sum of n 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

Scale factor = 1

6th term in the sequence = arn-1 = 1.26-1 = 32

3rd term form last = l/rn-1 = l/23-1 = 32/4 = 8

Sum of first 3 terms = a(rn -1)/(r – 1) = 1(23-1)/(2-1) = 7

### 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
the sequence is also in GP with the same common difference.
• Reciprocal of all the terms in GP also form 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.

## Explicit formula

An explicit formula is the one that defines terms of a sequence in relation to the term number. The nth term of a geometric sequence is given by the explicit formula:

an = a1 * rn-1

Problem: 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

### Recursive Formula

A recursive formula defines 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

Problem: 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

### Forms of geometric sequences for Conversions

Explicit form: an = k * rn-1

Recursive form: a1 = k , an = an-1 * r

Problem 1: Given recursive formula for f(n):

f(1) = 6

f(n) = f(n-1) * (-6.5)

Find an explicit formula for f(n)

Solution:

From the recursive formula, we can tell that the first term of the sequence is 6 and the common ratio is -6.5

Explicit formula : f(n) = 6 * (-6.5)n-1

Problem 2: Given explicit formula for f(n):

f(n) = 6 * (-6.5)n-1

Find recursive formula for f(n).

Solution:

From the explicit formula, we can tell that the first term of the sequence is 6 and the common ratio is -6.5

Recursive formula: f(1) = 6

f(n) = f(n-1) * (-6.5)

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