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

An arithmetic sequence is of the form:

\(a, ~a+d,~ a+2d,~â€¦â€¦~a~+~(n~-~2)d,~ a~+~(n~-~1)d\);

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 \(b,~ br, ~br^2~,~ â€¦â€¦~br^{n~-~2}, br^{n~-~1}\)

here \(b\) is the first term,

\(r\) is the common ratio

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

Let \(a_1,~ a_2, ~a_3,~ â€¦â€¦.~a_n\) be an A.P and \(b_1,~ b_2,~ b_3,~ â€¦â€¦.~b_n\) be a G.P

Then, \(a_1 ~b_1,~ a_2~ b_2,~ a_3~ b_3,~ â€¦â€¦, ~a_n~ b_n\) is called as an arithmetic geometric sequence and it is of the form,

\(ab,~ (a~+~d)br, ~(a~+~2d)br^2,~ â€¦â€¦.,~[a~+~(n-2)d]~br^{n~-~2}, ~[a~+~(n~-~1)d]br^{n~-~1}\)

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

\(S_n =ab+ (a+d)br +(a+2d)br^2+â€¦â€¦.+ [a+(n-2)d]br^{n-2}+ [a+(n-1)d]br^{n-1}\)

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

\(S_n =ab+ (a+d)br +(a+2d)br^2+â€¦â€¦.+ [a+(n-2)d]br^{n-2}+ [a+(n-1)d]br^{n-1}\]”> –â€“(1)

Now, multiplying each term with \(r\) gives,

\(rS_n = abr+(a+d)br^2+ (a+2d)br^3+â€¦.+[a+(n-2)d]br^{n-1}+[a+(n-1)d]br^n\) –â€“(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).

\(S_n~-~rS_n = ab~+~dbr~+~dbr^2~+~dbr^3~+~ â€¦â€¦~+ ~dbr^{n-1}~â€“~[a~+~(n~-~1)d]br^n\) —(3)

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

\(dbr~+~dbr^2~+~ dbr^3~ +~ â€¦â€¦â€¦.~+ ~dbr^{n~-~1}\) is a G.P

Sum of \(n\) terms of G.P is a \(\frac{1~-~r^n}{1~-~r}\),

Therefore, \(dbr~+~dbr^2~+~ dbr^3~ +~ â€¦â€¦â€¦.~+ ~dbr^{n~-~1}\) = \(dbr\frac{1~-~r^{n~-~1}}{1~-~r}\)

Now, (3) becomes as,

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

where \(r~â‰ ~0\)

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

\(Â \lim\limits_{nâ†’âˆž}~r^n\) = \(0\)

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 \(âˆšab\).

For example; Geometric mean of 3 and 27 is âˆš(3Ã—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} = \(4~Ã—~2\) = 8

G_{2} = \(8~Ã—~2\) = 16

G_{3} = \(16~Ã—~2\) = 32

**Relation between Arithmetic mean (A.M) and Geometric mean (G.M):**

Consider two positive numbers \(a\) and \(b\);

\(A.M\) = \(\frac{a~+~b}{2}\)

\(G.M\) = \(âˆšab\)

\(A.M~-~G.M\) = \(\frac{a~+~b}{2}~-~âˆšab\) = \(\frac{a~+~b~-~2âˆšab}{2}\)

\(A.M~-~G.M\) = \(\frac{(âˆša~-~âˆšb)^2}{2}\) which is greater than or equal to 0.

Therefore, \(A.M~â‰¥~G.M\)

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

Let the numbers be \(a\) and \(b\),

\(\frac{a~+~b}{2}\) = \(5\), \(a~+~b\) = \(5~Ã—~2\) = \(10\) Â Â Â Â —(1)

\(âˆšab\) = \(4\), \(ab\) = \(16\)

\((a~+~b)^2~-~(a~-~b)^2\) = \(4ab\)

\({10}^2~-~(a~-~b)^2\) = \(4~Ã—~16\) = \(64\)

\((a~-~b)^2\) = \(100~-~64\) = \(36\)

\(a~-~b\) = \(Â±6\) Â Â Â Â Â —(2)

By solving (1) and (2),

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

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

Thus here are some examples of the arithmetic geometric progression discussed above. Math enthusiasts who would like to practice more with sample questions and answers should visit the page provide here- NCERT solutions for arithmetic progressions and geometric progressions.