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

where \(r~≠~0\)

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

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 G_{1},G_{2},G_{3}……G_{n} be the n numbers inserted between the positive numbers a and b such that,

a,G_{1},G_{2},G_{3}……G_{n} 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, G_{1 = ar = \(a\left(\frac{b}{a}\right)^{\frac{1}{n~+~1}}\)}

G_{2 = ar2 = \(a\left(\frac{b}{a}\right)^{\frac{2}{n~+~1}}\)}

G_{2} = ar^{3} = \(a(\frac{b}{a})^{\frac{3}{n~+~1}}\)

G_{n} = ar^{n} = \(a(\frac{b}{a})^{\frac{n}{n~+~1}}\)

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

Let G_{1},G_{2},G_{3} be the three numbers to be inserted between 4 and 64,

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

G_{1} = \(4~×~2\) = 8

G_{2} = \(8~×~2\) = 16

G_{3} = \(16~×~2\) = 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),

\(2a\) = \(16\), \(a\) = \(\frac{16}{2}\) = \(8\)

\(b\) = \(10~-~a\) = \(10~-~8\) = \(2\)