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# Geometric Sequence

Top
 Sub Topics An ordered list of numbers is known as a sequence and the sum of the terms of a sequence is called a series. Some sequences have a pattern that is used to arrive at the sequence's terms. One such sequence is geometric sequence.A geometric sequence is a sequence in which the ratio of the consecutive terms will be a constant. It is said to be an exponential function and is written as a$_{n}$ = cr$^{n}$where r : Common ratioc : Constant (except the first term of the sequence)

## Definition

A sequence where each new term after the first is obtained by multiplying the preceding term by a constant r, called the common ratio is known as a geometric sequence.

If the first term in a sequence is a then the geometric  sequence is

a, ar, ar$^{2}$, ar$^{3}$,.......

N-th term of the above sequence is ar$^{n-1}$

Divide the second term by the first term to find the common ratio..

## Formula

Below are some important formulas used in geometric sequence:
Let given sequence is a, ar, ar$^{2}$, ar$^{3}$,....... ,where a is the first term and r is the common ratio. then

1) n${th}$ term of the sequence is a$_{n}$ = ar$^{n-1}$

2) Sum of certain number of terms, S$_{n}$ = $\frac{a(1-r^{n})}{1 - r}$ ; r < 1

or S$_{n}$ = $\frac{a(r^{n}-1)}{ r-1}$ ; r > 1

3) Sum of infinite terms: $\frac{a}{1-r}$; r $\in$ (-1, 1) and r $\neq$ 0

## Sum of Geometric Sequence

Sum of the first n terms of a geometric progression is:

S$_{n}$ = a + ar + ar$^{2}$ + ar$^{3}$ + .......... + ar$^{n-1}$

Multiply the above sum by r and subtract from previous eqaution

rS$_{n}$   = ar + ar$^{2}$ + ar$^{3}$ + .......... + ar$^{n-1}$ + ar$^{n}$

Now S$_{n}$ - rS$_{n}$  = a - ar$^{n}$

$\rightarrow$ S$_{n}$(1 - r) = a(1 - r$^{n}$)

Dividing by (1 - r) we get

S$_{n}$ = $\frac{a(1-r^{n})}{1 - r}$

The above formula can be used to find the sum of first n terms when the starting value is a and the common ratio is r.

## Examples

Some of the examples based on geometric sequence are as follows:

Example 1: Find the sum of 1 + 2 + 4 + 8 +..... to 8 terms.

Solution: Here a = 1, r = 2, n = 8

Let S = 1 + 2 + 4+ 8 +........... to 8 terms

$\frac{1(2^{8}- 1)}{2 - 1}$

= 2$^{8}$ - 1

= 255

Sum of 1 + 2 + 4 + 8 +..... to 8 terms is 255.

Example 2: Find the sum to n terms of the series
3 + 33 + 333 + ..........

Solution: Let S denote the required sum.

S = 3 + 33 + 333 +...... to n terms

= 3(1 + 11 + 111 +  ........ to n terms)

= $\frac{3}{9}$(9 + 99 + 999 + ...... to n terms)

= $\frac{3}{9}$ ( 10- 1) +(10$^2$ - 1) +(10$^3$ - 1) + ........ +(10$^n$ - 1)

= $\frac{3}{9}$ ((10 + 10$^{2}$ +10$^{3}$ +10$^n$) - n)

= $\frac{3}{9}$ (10(1 +10 +10$^{2}$ +.........+10$^ {n - 1}$) - n)

= $\frac{3}{9}$ (($\frac{10 (10^{n} - 1)}{10 - 1}$) - n)

= $\frac{3}{81}$ (10$^ {n + 1}$  - 10 - 9n)

= $\frac{1}{27}$(10$^{n+ 1}$ - 9n - 10)

Example 3: Find three numbers in geometric progression whose sum is 19 and the product is 216.

Solution: Let the three numbers be $\frac{a}{r}$ a, ar

$\frac{a}{r}$ * a * ar = 216.

a$^{3}$  = 6$^{3}$

a = 6

So the numbers are $\frac{6}{r}$, 6, 6r

$\frac{6}{r}$ + 6 + 6r = 19

6+ 6r$^{2}$ = 13r

6r$^{2}$ - 4r - 9r + 6 = 0

2r(3r - 2) - 3(3r - 2) = 2

(3r - 2) (2r - 3) = 0

r = $\frac{2}{3}$, $\frac{3}{2}$

So the numbers are

$\frac{6}{\frac{2}{3}}$, 6, 6*($\frac{2}{3}$) = 9, 6, 4

$\frac{6}{\frac{3}{2}}$, 6, 6*($\frac{3}{2}$) = 4, 6, 9