__Arithmetic Progression:__

__Arithmetic Progression:__

An **Arithmetic Progression (AP)** or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

For instance, the sequence 5, 7, 9, 11, 13, 15 … is an arithmetic progression with common difference of 2.

Its general form can be given as **a, a+d, a+2d, a+3d,…**

If the initial term of an arithmetic progression is *a* and the common difference of successive members is *d*, then the *n*th term of the sequence (a_{n}) is given by:

**a _{n} = a + (n – 1)d**

and in general

**Nth Term of A.P. is A _{n} = a_{m} + (n – m)d**

The sum of the members of a finite arithmetic progression is called an **arithmetic series **and given by,

**Sum of N terms of an A.P. is S _{n }= ^{n}/_{2} [2a + (n – 1)d] = ^{n}/_{2 }(a + l)**

**Arithmetic mean: **

When three quantities are in AP, the middle one is said to be the Arithmetic Mean (AM) of the other two, thus *a* is the AM of *(a-d)* and *(a+d).*

Arithmetic mean between two numbers a and b is given by,

**AM = (a+b)/ _{2}**

**Geometric progression:**

A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

For example, the sequence 2, 6, 18, 54, … is a geometric progression with common ratio 3.

The general form of a geometric sequence is *a, ar, ar ^{2},ar^{3},ar^{4},…*

A **geometric series** is the sum of the numbers in a geometric progression.

Let *a* be the first term and *r* be the common ratio, **a _{n}** nth term, n the number of terms, and

**S**be the sum up to n terms:

_{n}**The n-th term is given by, a_{n} = ar^{n-1}**

**The Sum up to n-th term of Geometric progression (G.P.) **is given by,

*If r > 1, then*

**S _{n }= a(r^{n}-1)/(r-1)**

*if r < 1, then*

**S _{n} = a(1-r^{n})/(1-r)**

Sum of infinite geometric progression when r<1:

**S _{n} = a/(1-r)**

**Geometric Mean (GM) **between two numbers a and b is given by,

**GM = sqrt ab **

__Some useful results on number series:__

**Sum of first n natural numbers is given by**

S = 1 + 2 + 3 + 4 +….+n

**S = n/2 * (n+1) **

**Sum of squares of the first n natural numbers is given by**

S = 1^{2} + 2^{2} + 3^{2} +….+n^{2}

**S = [{n(n+1)(2n+1)}/6 ]**

**Sum of cubes of the first n natural numbers is given by**

S = 1^{3} + 2^{3} + 3^{3} +….+n^{3}

**S = [{n(n+1)}/2]**

**Sum of first n odd natural numbers **

S = 1 + 3 + 5 +…+ (2n-1)

**S = n ^{2}**

**Sum of first n even natural numbers S = 2 + 4 + 6 +…+ 2n**

**S = n(n+1)**

*Note:*

1) If we are counting from n1 to n2 including both the end points, we get **(n2-n1) + 1** numbers.

e.g. between 12 and 22, there is (22-12) +1 = 11 numbers (Including both the ends).

2) In the first n, natural numbers:

i) If n is **even**

There are **n/2** odd and **n/2** even numbers

e.g from 1 to 40 there are 25 odd numbers and 25 even numbers.

ii) If n is odd

There are **(n+1)/2** odd numbers, and **(n-1)/2** even numbers

e.g. from 1 to 41, there are (41+1)/2= 21 odd numbers and (41-1)/2 = 20 even numbers.

### Harmonic Progression (HP)

A sequence of numbers is called a harmonic progression if the reciprocal of the terms are in AP. In simple terms, a,b,c,d,e,f are in HP if 1/a, 1/b, 1/c, 1/d, 1/e, 1/f are in AP.

- Harmonic Mean = (2 a b) / (a + b)

For two numbers, if A, G and H are respectively the arithmetic, geometric and harmonic means, then

- A ≥ G ≥ H
- A H = G
^{2}, i.e., A, G, H are in GP

**Arithmetico-Geometric series**

A series having terms a, (a+d)r, (a+2d)r^{2},…… etc is an Arithmetico-Geometric series where a is the first term, d is the common difference of the Arithmetic part of the series and r is the common ratio of the Geometric part of the series.

The nth term tn= [a+(n-1)d]r^{n-1}

The sum of the series to n terms is

Sn= a/1-r+ dr (1-r^{n-1})/ (1-r)^{2} – [a+(n-1)d]rn/ 1-r

The sum to infinity, S= a/ 1-r + dr (1-r^{n-1})/(1-r)^{2} ; r<1

**Exponential Series**

e^{x }= 1+x/1!+x^{2}/2!+x^{3}/3!+………. (e is an irrational number)

coefficient of xn= 1/n!; Tn+1=x^{x}/n!

e^{-x} = 1-x/1!+x2/2!- x3/3!+……….

**Logarithmic Series**

log_{e} (1+x)= x-x^{2}/2+x^{3}/3+x^{4}/4+…….. (-1<x 1)

log_{e }(1-x)= -x-x^{2}/2-x^{3}/3- x^{4}/4 -…….. (-1x< 1)

log_{e} (1+x)/(1-x)= 2-(x+x^{3}/3+x^{5}/5+………) (-1<x 1)

### Sample Problems

**Question 1 : **Find the nth term for the AP : 11, 17, 23, 29, …**Solution : **Here, a = 11, d = 17 – 11 = 23 – 17 = 29 – 23 = 6

We know that nth term of an AP is a + (n – 1) d

=> nth term for the given AP = 11 + (n – 1) 6

=> nth term for the given AP = 5 + 6 n

We can verify the answer by putting values of ‘n’.

=> n = 1 -> First term = 5 + 6 = 11

=> n = 2 -> Second term = 5 + 12 = 17

=> n = 3 -> Third term = 5 + 18 = 23

and so on …

**Question 2 : **Find the sum of the AP in the above question till first 10 terms.**Solution : **From the above question,

=> nth term for the given AP = 5 + 6 n

=> First term = 5 + 6 = 11

=> Tenth term = 5 + 60 = 65

=> Sum of 10 terms of the AP = 0.5 n (first term + last term) = 0.5 x 10 (11 + 65)

=> Sum of 10 terms of the AP = 5 x 76 = 380

**Question 3 : **For the elements 4 and 6, verify that A ≥ G ≥ H.**Solution : **A = Arithmetic Mean = (4 + 6) / 2 = 5

G = Geometric Mean = = 4.8989

H = Harmonic Mean = (2 x 4 x 6) / (4 + 6) = 48 / 10 = 4.8

Therefore, A ≥ G ≥ H

**Question 4 : **Find the sum of the series 32, 16, 8, 4, … upto infinity.**Solution : **First term, a = 32

Common ratio, r = 16 / 32 = 8 / 16 = 4 / 8 = 1 / 2 = 0.5

We know that for an infinite GP, Sum of terms = a / (1 – r)

=> Sum of terms of the GP = 32 / (1 – 0.5) = 32 / 0.5 = 64

**Question 5 : **The sum of three numbers in a GP is 26 and their product is 216. ind the numbers.**Solution : **Let the numbers be a/r, a, ar.

=> (a / r) + a + a r = 26

=> a (1 + r + r^{2}) / r = 26

Also, it is given that product = 216

=> (a / r) x (a) x (a r) = 216

=> a^{3} = 216

=> a = 6

=> 6 (1 + r + r^{2}) / r = 26

=> (1 + r + r^{2}) / r = 26 / 6 = 13 / 3

=> 3 + 3 r + 3 r^{2} = 13 r

=> 3 r^{2} – 10 r + 3 = 0

=> (r – 3) (r – (1 / 3) ) = 0

=> r = 3 or r = 1 / 3

Thus, the required numbers are 2, 6 and 18.