# Arithmetic Geometric Sequence

Arithmetic Geometric sequence is the fusion of an arithmetic sequence and a geometric 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.

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.

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:

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

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)

Write first term, second term, third term of (2) below the second term, third term, fourth term of (1) respectively and so on and subtract (2) from (1).

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

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

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

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, $A.M~≥~G.M$

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

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),

$2a$ = $16$, $a$ = $\frac{16}{2}$ = $8$

$b$ = $10~-~a$ = $10~-~8$ = $2$