`\color{fuchsia} \(★ mathbf\ul"Geometric Progression (G . P.)")`

A sequence `color(blue)(a_1, a_2, a_3, …, a_n, … )` is called geometric progression,

if each term is non-zero and `color(red)(a_(k+1)/a_k = r)` (constant ratio ), for `color(red)(k ≥ 1.)`

`\color{red} ✍️` By letting `color(blue)(a_1 = a)`, we obtain a geometric progression, `color(red)(a, ar, ar^2, ar^3,…...,)`

where `a` is called the `color(blue)"first term"` and `r` is called the `color(blue)"common ratio"` of the G.P.

We shall use the following notations with these formulae:

`color(red)(a) = ` the first term, ` \ \ color(red)(r) = `the common ratio, ` \ \ color(red)(l) =` the last term,

`color(red)(n) =` the numbers of terms,

`color(red)(S_n) =` the sum of n terms.

`color(red)("Let us consider the following sequences:")`

`(i) color(blue)(2,4,8,16,.............,) (ii) color(blue)(1/9 , -1/ 27, 1/81 , -1/243 ..............) (iii) color(blue)(.01,.0001,.000001,..................)`

In `(i)` we have `a_1=2 , a_2/a_1=2, a_3/a_2=2, a_4/a_3=2` and so on. In `(ii)`, we observe, `a_1=1/9 , a_2/a_1=-1/3, a_3/a_2=-1/3 , a_4/a_3=-1/3` and so on.

Similarly, state how do the terms in `(iii)` progress

In `(i),` this constant ratio is `2;` in `(ii)` it is `-1/3` and in `(iii)` the constant ratio is `0.01.`

Common ratio in geometric progression `(i), (ii)` and `(iii)` above are `2, -1/3` and `0.01,` respectively.

Such sequences are called geometric sequence or geometric progression abbreviated as `G.P`

Let us consider a `G.P. `with first non-zero term `‘a’` and common ratio `‘r’.`

The second term is obtained by multiplying `a` by `r,` thus `color(blue)(a_2 = ar).`

Similarly, third term is obtained by multiplying `a_2` by `r.` Thus, `color(blue)(a_3 = a_2r = ar^2),` and so on.

We write below these and few more terms.
1st term `color(blue)(= a_1 = a = ar^(1–1)),`

2nd term `color(blue)(= a_2 = ar = ar^(2–1)),`

3rd term `color(blue)(= a_3 = ar^2 = ar^(3–1))`,

4th term `color(blue)(= a_4 = ar^3 = ar^(4–1)),`

5th term `color(blue)(= a_5 = ar^4 = ar^(5–1))`,

Do you see a pattern? What will be `16^(th)` term?

`color(blue)(a_16 = ar^(16–1) = ar^15)`

Therefore, the pattern suggests that the `n^(th)` term of a G.P. is given by `color(red)(a _n =ar^( n−1).)`

Thus, a, G.P. can be written as `color(blue)(a, ar, ar^2, ar^3, … ar^(n – 1)) \ \ \ \ \; color(red)(a, ar, ar^2,...,ar^(n – 1)....... ;)` according as G.P. is finite or infinite, respectively.

The series `color(blue)(a + ar + ar^2 + ... + ar^(n–1))` or `color(red)(a + ar + ar^2 + ... + ar^(n–1) +....... )` are called finite or infinite geometric series, respectively.

Let the first term of a G.P. be `a` and the common ratio be `r.`

Let us denote by `S_n` the sum to first `n` terms of `G.P.` Then

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(blue)(S_n = a + ar + ar^2 +...+ ar^(n–1)) .................... (1)`

`color(red)(Case 1)` If `color(blue)(r = 1),` we have `color(blue)(S_n = a + a + a + ... + a)` (n terms) `color(red)(= na)`

`color(red)(Case 2)` If `color(blue)(r ≠ 1),` multiplying (1) by `r,` we have

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(blue)(rS_n = ar + ar^2 + ar^3 + ... + ar^n) ........................ (2)`

Subtracting (2) from (1), we get `(1 – r) S_n = a – ar^n = a(1 – r^n)`

`color(red)(S_n= (a(1 – r^n))/(1-r))` or `color(blue)(S_n= (a(r^n-1))/(r-1))`

`\color{fuchsia} \(★ mathbf\ul"Geometric Mean (G .M.) ")`

The geometric mean of two positive numbers `a` and `b` is the number `color(red)(sqrt(ab )).`

Therefore, the geometric mean of `2` and `8` is `4.`

We observe that the three numbers `2,4,8` are consecutive terms of a G.P. This leads to a generalisation of the concept of geometric means of two numbers.

Given any two positive numbers a and b, we can insert as many numbers as we like between them to make the resulting sequence in a G.P.

Let `color(blue)(G_1, G_2,…, G_n)` be `n` numbers between positive numbers `a` and `b` such that `color(blue)(a,G_1,G_2,G_3,…,G_n,b)` is a G.P.

Thus, `b` being the `(n + 2)th` term,we have `color(blue)(b = ar^(n + 1))` , or `color(blue)(r=(b/a)^(1/(n+1)))`

Hence `color(blue)(G_1= ar=a(b/a)^(1/(n+1)))`

`color(blue)(G_2= ar^2=a(b/a)^(2/(n+1)))`

`color(blue)(G_3= ar^3=a(b/a)^(3/(n+1)))`

`color(blue)(G_n=ar^n=a(b/a)^(n/(n+1)))`

Let `A` and `G` be A.M. and G.M. of two given positive real numbers `a` and `b,` respectively.

Then `color(blue)(A= (a+b)/2)` and `color(blue)(G=sqrt(ab))`

Thus, we have `color(blue)(A – G)=(a+b)/2-sqrt(ab)= (a+b-2sqrt(ab))/2`

` \ \ \ \ \ \ \ \ \ \ \ color(blue)(A – G)=(sqrt(a)-sqrt(b))^2/2 >= 0`

we obtain the relationship `color(red)(A ≥ G.)`