Geometric progression

Statement: A progression in which the ratio of two consecutive terms is always same constant, such a progression is called geometric progression or simply called GP. and the constant ratio is called common ratio

i.e. if $a_1,a_2,a_3,…….a_n$ are in GP then

$$\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\frac{a_5}{a_4}=………$$

General form of GP is $a, ar, ar^2, ar^3, ar^4 ……….$ where

first term is $a_1=a$

second term is $a_2=ar$

third term is $a_3=ar^2$ ……..

$n^{th}$ term is $a_n=ar^{n-1}$

Sum of n terms of a GP

Statement:

1. sum of first n terms of a geometric progression is $S_n= \frac{a(r^n-1)}{r-1}$ when r $\neq$ 1
2. if r=1 then $S_n$ = a + a + a + a + a + a +. . . . (n terms) = na
3. if |r|<1 and n = $\infty$ then $S_n = \frac{a}{1-r}$

Proof: Case-1)

$S_n= a+ar+ar^2+…..+ar^{n-1}$—-(1)

now multiply common ration r on both sides of the equation

$rS_n =ar+ar^2+ar^3+……+ar^n$—–(2)

let’s subtract equation (1) from equation (2)

$S_n = a+ar+ar^2+…..+ar^{n-1}$

$\underline{- rS_n = -(ar+ar^2+ar^3+……+ar^n) } $

$(1- r) S_n= [a- ar^n]$

$S_n= \frac{a(1- r^n)}{1-r}$

Case-3)

given |r|<1 and n = $\infty$

from the above proof we can write

$$ S_n= \frac{a(1- r^n)}{1-r}$$

as |r|<1 and n = $\infty$ then $r^n = 0$

$$ S_n= \frac{a(1- 0)}{1-r}$$

$$ S_n= \frac{a}{1-r}$$

Geometric mean (GM):

If $a_1,a_2,a_3,…….a_n$ are positive numbers then $(a_1a_2a_3…….a_n)^{1/n}$ is called geometric mean of $a_1,a_2,a_3,…….a_n$.

therefore GM of a and b is

$$GM=\sqrt{ab}$$