# Arithmetic Geometric Sequence

## Trigonometry # Arithmetic Geometric Sequence Arithmetic Geometric sequence is the fusion of an arithmetic sequence and a geometric sequence. In this article, we are going to discuss the arithmetic-geometric sequences and the relationship between them. Also, get the brief notes on the geometric mean and arithmetic mean with more examples.

## What is Arithmetic Sequence?

An arithmetic sequence is of the form:

$$a, ~a+d,~ a+2d,~……~a~+~(n~-~2)d,~ a~+~(n~-~1)d$$;

where $$a$$ is the first term,

$$d$$ is the common difference

$$a~+~(n~-~1)d$$ is $$n^{th}$$ term of A.P.

## What is a Geometric Sequence?

Similarly, a geometric sequence is of the form $$b,~ br, ~br^2~,~ ……~br^{n~-~2}, br^{n~-~1}$$

here $$b$$ is the first term,

$$r$$ is the common ratio

$$br^{n~-~1}$$ is $$n^{th}$$ term of the G.P.

## What is Arithmetic Geometric Sequence?

Let $$a_1,~ a_2, ~a_3,~ …….~a_n$$ be an A.P and $$b_1,~ b_2,~ b_3,~ …….~b_n$$ be a G.P
Then, $$a_1 ~b_1,~ a_2~ b_2,~ a_3~ b_3,~ ……, ~a_n~ b_n$$ is called as an arithmetic geometric sequence and it is of the form,
$$ab,~ (a~+~d)br, ~(a~+~2d)br^2,~ …….,~[a~+~(n-2)d]~br^{n~-~2}, ~[a~+~(n~-~1)d]br^{n~-~1}$$

Sum of n terms of the above sequence is found as follows:

$$S_n =ab+ (a+d)br +(a+2d)br^2+…….+ [a+(n-2)d]br^{n-2}+ [a+(n-1)d]br^{n-1}$$ —- (1)

Now, multiplying each term with $$r$$ gives,

$$rS_n = abr+(a+d)br^2+ (a+2d)br^3+….+[a+(n-2)d]br^{n-1}+[a+(n-1)d]br^n$$ ––(2)

Subtract equation (2) from equation (1), we get

$$S_n~-~rS_n = ab~+~dbr~+~dbr^2~+~dbr^3~+~ ……~+ ~dbr^{n-1}~–~[a~+~(n~-~1)d]br^n$$ —(3)

In the above series, if we exclude the first term and the last term,

$$dbr~+~dbr^2~+~ dbr^3~ +~ ……….~+ ~dbr^{n~-~1}$$ is a G.P
Sum of $$n$$ terms of G.P is a $$\frac{1~-~r^n}{1~-~r}$$,
Therefore, $$dbr~+~dbr^2~+~ dbr^3~ +~ ……….~+ ~dbr^{n~-~1}$$ = $$dbr\frac{1~-~r^{n~-~1}}{1~-~r}$$

Now, (3) becomes as,

(1-r) Sn = $$ab ~+~ dbr\frac{1~-~r^{n~-~1}}{1~-~r}~-~[a~+~(n~-~1)d]br^n$$,

where $$r~≠~0$$

Sn = $$\frac{ab}{1~-~r}~+~dbr\frac{1~-~r^{n~-~1}}{(1-r)^2}~ -~\frac{[a~+~(n-1)d]br^n}{1~-~r}$$

If the common ratio of the sequence lies between -1 and 1, then

$$\lim\limits_{n→∞}~r^n$$ = $$0$$

Therefore, the sum of infinite terms of the sequence in (1) is,

S = $$\frac{ab}{1~-~r} ~+ ~\frac{dbr}{(1-r)^2}$$ [-1<r<1]

## Geometric Mean

Consider two positive numbers a and b, the geometric mean of these two numbers is $$√ab$$.

For example; Geometric mean of 3 and 27 is √(3×27)=9

The numbers 3, 9, 27 is in a G.P with common ratio 3.

In general; between 2 positive numbers a and b, we can insert as many numbers as we like such that the resulting sequence forms a G.P.

Let G1, G2, G3……Gn be the n numbers inserted between the positive numbers a and b such that,

a,G1,G2,G3……Gn forms a G.P. “a” is the first term and b is (n+2)th term.

b = ar{n+1},

$$r = \left(\frac{b}{a}\right)^{\frac{1}{n~+~1}}$$

It gives, G1 = ar = $$a\left(\frac{b}{a}\right)^{\frac{1}{n~+~1}}$$
G2 = ar2 = $$a\left(\frac{b}{a}\right)^{\frac{2}{n~+~1}}$$
G2 = ar3 = $$a(\frac{b}{a})^{\frac{3}{n~+~1}}$$
Gn = arn = $$a(\frac{b}{a})^{\frac{n}{n~+~1}}$$

### Relation Between Arithmetic Mean (A.M) and Geometric Mean (G.M)

Consider two positive numbers $$a$$ and $$b$$;
$$A.M$$ = $$\frac{a~+~b}{2}$$
$$G.M$$ = $$√ab$$
$$A.M~-~G.M$$ = $$\frac{a~+~b}{2}~-~√ab$$ = $$\frac{a~+~b~-~2√ab}{2}$$
$$A.M~-~G.M$$ = $$\frac{(√a~-~√b)^2}{2}$$ which is greater than or equal to 0.

Therefore, $$A.M~≥~G.M$$

### Arithmetic Geometric Sequence Example

Example 1:

Insert 3 numbers between 4 and 64 so that the resulting sequence forms a G.P.

Solution:

Let G1,G2,G3 be the three numbers to be inserted between 4 and 64,

$$r = (\frac{b}{a})^{\frac{1}{n~+~1}} = (\frac{64}{4})^{\frac{1}{3~+~1}} = (16)^{\frac{1}{4}} = 2$$

G1 = $$4~×~2$$ = 8

G2 = $$8~×~2$$ = 16

G3 = $$16~×~2$$ = 32

Example 2:

A.M and G.M of two numbers are 5 and 4 respectively. Find the two numbers.

Solution:

Let the numbers be $$a$$ and $$b$$,
$$\frac{a~+~b}{2}$$ = $$5$$, $$a~+~b$$ = $$5~×~2$$ = $$10$$         —(1)
$$√ab$$ = $$4$$, $$ab$$ = $$16$$
$$(a~+~b)^2~-~(a~-~b)^2$$ = $$4ab$$
$${10}^2~-~(a~-~b)^2$$ = $$4~×~16$$ = $$64$$
$$(a~-~b)^2$$ = $$100~-~64$$ = $$36$$
$$a~-~b$$ = $$±6$$           —(2)

By solving (1) and (2),

$$2a$$ = $$16$$, $$a$$ = $$\frac{16}{2}$$ = $$8$$
$$b$$ = $$10~-~a$$ = $$10~-~8$$ = $$2$$

Thus, some examples of the arithmetic geometric progression are discussed above. To learn different kinds of progressions in Maths, download BYJU’S – The Learning App.