progression

• 1: Arithmetic progression
• 2: Geometric progression
• 3: Relationship between Arithematic and Geometric Means

1 - Arithmetic progression

A progression in which the difference between any two consecutive terms his always as same fixed quantity, such a progression is called arithmetic progression or simply AP. And the fixed difference is called common difference.

If a is the first time and t is the common difference then general AP form is:

a, a+t, a+2t, a+3t ……..

Where we denote terms as follows:

$a_1$ = a

$a_2$ = a+t

$a_3$ = a+ 2t and so on.

General term $a_n$ = a+(n-1)t . That is $a_n$ is the $n^th$ term with a being the first term and t is the common difference.

Examples of AP

1. 1,2,3,4,5,6,7……..

2. 1,3,5,7,9….

3. 9,6,3,0,-3….

4. 1,11,21,31,….

Sum of first n terms of AP

statment: Let $S_n$ be the sum of first n terms of AP $a_1,a_2,a_3,…….a_n$ then $S_n$ = $\frac{n}{2}[2a+(n-1)t]$ = $\frac{n}{2}[a_1+a_n]$

Proof:

Given the AP, $a_1,a_2,a_3,…….a_n$

$$S_n = a_1 + a_2 + a_3 + …… + a_n$$

$$S_n = a + (a+t) + (a+2t) + ….. + [a+ (n-1)t]$$

$$S_n = \sum_{r=1}^n[a+(n-1)t]$$

$$S_n=\sum_{r=1}^{n} a + \sum_{r=1}^{n}(n-1)t $$

As a and t are constants we can take them out of summation

$$S_n = a \sum_{r=1}^{n} 1 + t \sum_{r=1}^{n} (n-1)$$

$$Here \sum_{r=1}^{n} 1 = n \ because\ 1\ is\ added\ n\ times$$

$$S_n = a n + t [\sum_{r=1}^{n} n - \sum_{r=1}^{n} 1]$$

$$And\ we\ know\ that\ sum\ of\ first\ n\ natural\ numbers\ i.e. \sum_{r=1}^{n} n = \frac{n(n+1)}{2}$$

$$S_n = a n + t [ \frac{n(n+1)}{2} - n]$$

$$S_n = n [ a + t (\frac{(n+1)}{2} -1)]$$

$$S_n = n [ a + t \frac{(n-1)}{2} ]$$

$$S_n = \frac{n}{2} [ 2a + t (n-1)]$$

Points to be noted

1. If every term in an AP is multiplied by a constant k then new series will also be in AP with common difference kt
2. If every term in an AP is added with k then the new series will be in AP with common difference (k+t)

Arithmetic Mean (AM)

Definition: If $a_1, a_2, a_3, …..a_n$ are real numbers then $\frac{a_1 + a_2 + a_3 + ….. + a_n}{n}$ is called arithmetic mean of $a_1, a_2, a_3, …..a_n$.

In such AM we can observe something interesting. AM of any two numbers make another AP

i.e. if A is the AM of a and b then a, A, b are in AP

If b is the AM of a and c then a, b, c are in AP

Proof:

Given b is the arithematic mean of a and c that means

b = $\frac{a+c}{2}$

for a, b ,c to be in AP the difference between adjacent terms must be equal i.e.

b-a = c - b

as b = $\frac{a+c}{2}$

b-a = $\frac{a+c}{2}$ - a = $\frac{c-a}{2}$

and c-b = c - $\frac{a+c}{2}$ = $\frac{c-a}{2}$

we got b-a = c-b

therefore a,b,c are in AP.

2 - 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

$$GM=\sqrt{ab}$$

3 - Relationship between Arithematic and Geometric Means

Relation: For A being the arithematic mean and G being the Geometric mean of positive real numbers a and b then we habe A= $\frac{a+b}{2}$ and G= $\sqrt {ab}$ then $$A \geq G$$

Proof: Given two positive numbers a and b and A and G are arithematic and geometric means respectively.

$$A-G=\frac{a+b}{2}-\sqrt {ab}$$

$$A-G=\frac{a+b-2\sqrt {ab}}{2}$$

$$A-G=\frac{(\sqrt a)^2+(\sqrt b)^2-2\sqrt {ab}}{2}$$

using the algebraic formula $a^2+b^2-2ab=(a+b)^2$ in the equation

$$A-G=\frac{(\sqrt a-\sqrt b)^2}{2}$$

for two positive numbers a and b $(\sqrt a-\sqrt b)^2$ is always positve or zero, which can be represented as

$$A-G=\frac{(\sqrt a-\sqrt b)^2}{2} \geq 0$$

that implies

$$A \geq G$$

hence proved for two numbers

here $A = G$ would mean that a=b. that is for the same number both AM and GM will be same.

Similarly this can be extrapolated and can be expressed more generally as for $a_1, a_2, a_3, …..a_n$ where all being positive numbers then arithematic mean A = $\frac{a_1 + a_2 + a_3 + ….. + a_n}{n}$ and geometric mean $(a_1a_2a_3…….a_n)^{1/n}$. Then it can be shown that

$$A \geq G$$

and A=G when $a_1= a_2= a_3= …..=a_n$