# Geometric Sequence and Series

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:

a

_{1}, a_{2}, a_{3,}a_{4}, …, a_{n}

Given sequence is a geometric sequence if:

a

_{1}/a_{2}= a_{2}/a_{3}= a_{3}/a_{4}= … = a_{n-1}/a_{n }= 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

Common ratio is given by:

r = successive term/preceding term = ar

^{n-1 }/ ar^{n-2}

## What is the N^{th} 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, ar^{2}, ar^{3}, ar^{4}……….

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_{1}a

_{2 }= a_{1}ra

_{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}…

a

_{n }= a_{1}r^{n – 1}

**where, **

a

_{1}= the first term, a_{2}= the second term, and so on

athe last term (or the nth term) and_{n}:

aany term before the last term_{m}:

**nth term from the last term is given by:**

a

_{n}= l/r^{n-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:

S

_{n}= a(1 – r^{n})/(1 – r), if r < 1S

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

### 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 – 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**

**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 = ar

^{n-1 }= 1.2^{6-1}= 323rd term form last = l/r

^{n-1}= l/2^{3-1 }= 32/4 = 8Sum of first 3 terms = a(r

^{n}-1)/(r – 1) = 1(2^{3}-1)/(2-1) = 7

### 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

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 y
^{2}= 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:

**a _{n }= a_{1 }* r^{n-1}**

**Problem: 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

### 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 a

_{1}term(n) = term(n – 1) * r

= term(n -1) * 1.5 for n>=2

a

_{1 }= 6

**Forms of geometric sequences for Conversions**

Explicit form:a_{n}= k * r^{n-1}

Recursive form:a_{1 }= k , a_{n }= a_{n-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)