# Geometric Sequence and Series

Euclid’s book The Elements (300 BC, Book VIII) introduces a “geometric progression” as a progression in which the ratio of any element to the previous element is a constant.
The Greeks, over two thousand years ago, considered sequences such as
$\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},....,\frac{1}{2^k},...$
and their sums, such as
$\displaystyle \sum_{k=1}^{k=\infty} \frac{1}{2^k}$ The sum, above, at any finite step, is always less than the number 1. Since the sum is less than 1 at any finite step , we can conclude that the series converges in the limit to 1

Does a series have a sum?
$\sum_{n=1}^\infty \frac{1}{2^n}$
$\sum_{n=1}^\infty \frac{1}{2^n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...=1$
The sum of the series is 1.
The visualization is as follows
Imagine that we paint a blank canvas in steps. At each step, we paint half of the unpainted area. The total area painted after "n" steps is therefore the "n"th partial sum
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+ \frac{1}{2^n}$
The total area remaining unpainted is
$\frac{1}{2^n}$
After an infinite number of steps we will have painted all of the canvas, of which the area is 1.

#### Geometric sequence

An geometric sequence is a list of numbers with a definite pattern. If you take any number in the sequence then divide it by the previous one, and the result is always the same or constant then it is an geometric sequence.

In the geometric sequence,
The constant ratio in all pairs of consecutive or successive numbers in a sequence is called the common ratio. It is denoted by a letter r.
Ratio here means the second divide the first. We use the common ratio to go from one term to another.

Therefore, a geometric progression (GP) or geometric sequence is a sequence of numbers such that the ratio between the consecutive terms is always constant. For instance,
the sequence 5, 15, 45,135, . . . is an geometric progression with common ratio of 3.

#### General term of geometric sequence

A geometric sequence with a starting value a and common ratio r is a sequence of the form
$a,ar,ar^2,ar^3,...$
A recursive definition for this sequence has two parts
$a_1=a:$ initial condition
an=$a_{n-1}r$ for n>1 recursive formula
Therefore, an explicit definition for this sequence is a single formula as general terms given by
an=$ar^{n-1}$for n > 1

#### Example 1

When a ball bounces as shown below, the height of consecutive bounces become a geometric sequence. What is the height of 5th bounce?

The height of the first and third bounces are given in the figure above.

Solution
$a_1=100,a_3=49$
Use the explicit formula to relate a1 to a3 and to find r.
or an=$a_1 r^{n-1}$
or $a_3=a_1 r^{3-1}$
or $49=100 r^2$
or $r=\frac{7}{10}$
Find the fifth using explicit formula, thus we have
$a_5=a_1 r^4$
or $a_5=100 \times \left ( \frac{7}{10} \right ) ^4 \approx 24$

#### Geometric mean

In the geometric progression (GP), if a, b and c are three consecutive terms, then b is called geometric mean of a and b.
In this sequence, the first two terms a and b will have the ratio which will be equal to the next two terms b and c.
So we can say, b/a = c/b.
Rearranging the terms, we get
$b^2 = a \times c$
or $b = \sqrt{ac}$

#### Example 2

[Integration to Geometry]: The illustration below shows a sequence of $30^0,60^0,90^0$ triangles. Draw the next triangle in the sequence.

#### Geometric Series

Let a geometric sequence is given by
$a,ar,ar^2,ar^3,...$
If we are adding up the terms of the geometric sequence given above, then we have a geometric sum or geometric series given by
$\displaystyle \sum_{k=0}^{k=\infty} ar^k$

#### Method 1

The explicit formula for the sum of finite terms is given as below
Sum of geometric sequence up to n th term: $a,ar,ar^2,ar^3,...ar^{n-1}$ is written as
Sn=$a+(a r)+(a r^2)+(a r^3)+---+[a r^{n-2}]+[a r^{n-1}]$
Taking a common, we get
Sn=$a[1+r+r^2+r^3+---+r^{n-2}+r^{n-1}]$
Dividing by 1-r , we get
Sn=$a \frac{1-r}{1-r} [1+r+r^2+r^3+---+r^{n-2}+r^{n-1}]$
or$S_n=\frac{a(r^n-1)}{r-1}$

#### Method 2

Sum of geometric sequence up to n th term: $a,ar,ar^2,ar^3,...ar^{n-1}$ is written as
Sn=$a+(a r)+(a r^2)+(a r^3)+---+[a r^{n-2}]+[a r^{n-1}]$
Multiplying by r, we get
rSn=$(a r)+(a r^2)+(a r^3)+---+[a r^{n-1}]+[a r^n]$
Substracting , we get
rSn-Sn=$ar^n-a$
or$S_n=\frac{a(r^n-1)}{r-1}$

#### Infinite Sum

We have seen that the sum of the first n terms of a geometric series with first term a and common ratio r is
$S_n=\frac{a(r^n-1)}{r-1} ; r \ne 1$
In the case when r has magnitude less than 1, the term $r^n$ approaches 0 as n becomes very large. So, in this case, the sequence of partial sums S1,S2,S3,... has a limit:
$S_\infty= \displaystyle \lim_{n \to \infty} \frac{a(r^n-1)}{r-1} = \frac{a}{1-r}$
The limiting sum is usually referred to as the sum to infinity of the series and denoted by S∞. Thus, for a geometric series with common ratio r such that |r | < 1, we have
$S_\infty= \frac{a}{1-r}$

#### Example 3

A person saves NRs100 in a bank account at the beginning of each month. The bank offers a return of 12% compounded monthly.
(a) Determine the total amount saved after 12 months.
(b) After how many months does the amount saved first exceed NRs2000?
Solution
We need explicit formula for the sum of geometric series.
Sn: Sum of geometric sequence up to n th term.
It is written as
Sn=$\frac{a(1-r^n)}{1-r}$
Based on the formula, the solution of above problem with a = 100(1.01), r = 1.01 and n = 12, is
$S_{12}=\frac{a(r^n-1)}{r-1}= 1280.93$

#### Example 4

What is the sum of the finite geometric series 3+6+12+24+...+3072
Given that, the first term is 3, the common ratio is 2, and the nth term is 3072, therefore, we use explicit formula to find the n

an=$a_1 r^{n-1}$
or $3072=3 \times 2^{n-1}$
or $1024=\times 2^{n-1}$
or $n=11$
Therefore, the sum up to 11th term is
or Sn=$\frac{a(1-r^n)}{1-r}$
or $S_{11}= \frac{3(1-2^{11})}{1-2}$
or $S_{11}= 6141$