5. Arithmetic Progressions-Introduction of Arithmetic Progression Mathematics Class 10 In English-CBSE Notes

Introduction of Arithmetic Progression

Sequence: A set of numbers arranged in some definite order and formed
according to some rules is called a sequence.
 

Progression: The sequence that follows a certain pattern is called
progression.

Arithmetic Progression: An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.  

Examples:

(a) 1, 2, 3, 4, 5 ...................................... 

(b)  2, 5, 8, 11, 14 ..................................

(c)  50, 45, 40, 35 .....................................

(d)  -10, -4, 2, 8 .........................................

In above examples, each of the numbers in the list is called a term. Each term is obtained by adding a fixed number except the first term. And we can write other next term by adding same fixed number. This fixed number is called the common difference of the AP. 

In other words, A list of numbers which has same common difference will be said Arithmetic progression. 

• Each number of an arithmetic progression is called term. 
• The first number of A.P is called first term.
• The last term of A.P is called final term. 
• Difference between two consicutive terms of an A.P is known as common difference. 

​The first term is denoted by a or a_1. and

Second term by a₂

Third term by a₃

Fourth term by a₄

and so on.

Common difference by d

the number of term is denoted by n

Final term is denoted by a_n

Important points and Formula: 

(A) General form of an A.P is 

a, a + d, a + 2d, a + 3d, a + 4d ............................................................

(where a is the first term and d is common difference)

so the,

(B) General term for n^th term is a_n = {a + (n -1)d}

(C) If a, b, and c are in A.P then 

This is also Known as Arithmetic Mean. 

(D)  If three numbers are in A.P then take a - d, a, a + d as general term. 

(E)  General formula for the sum of natural numbers begin with 1 like

       1 + 2 + 3 .......

(F) Sum of n^th  of an A. P:

(G)  If ‘l’ is the last term of a finite A.P., then the sum is given by

(H) If S_n is given, then n^th term is given by a_n = s_n – s_{n -1}.      

General formula for common difference:

d = a_n - a_{n -1}

Like,

d = a₂ - a₁

d = a₃ - a₂

d = a₅ - a₄

and so on ...

1. Checking for A. P 

Question 1:  Is 2, 4, 8, 16, . . . an A.P?

Solution:

d = a₂ - a₁ = 4 - 2 =  2

d = a₃ - a₁ = 8 - 4 = 4,

d = a₄ - a₃ = 16 - 8 = 8,

Here all the value of Common difference (d) is not a fixed number or common, these all are different like 2, 4 and 8.

Hence 2, 4, 8, 16 .............. is not an A.P.

Question2: Is – 10, – 6, – 2, 2, . . . an A. P?

Solution:

d = a₂ - a₁ = - 6 - (-10) = - 6 +10 = 4,

d = a₃ - a₁ = - 2 - ( -6) = - 2 + 6 = 4,

d = a₄ - a₃ = 2 - (-2) = 2 + 2 = 4

 Here all the value of common difference (d) is fixed or equal.

Hence – 10, – 6, – 2, 2, . . . is an A.P.

Writing Next terms of the A.P`

Question 1: Write the next four terms of the A.P, when the first term a and the common difference d are given as follows:

(i)  a = 4, d = -3                                  (ii)  a = 7, d = 13

Solution: 

(i)  general form of an A.P is 

      a, a + d, a + 2d, a + 3d ..............

      4, 4 + (-3), 4 + 2(-3), 4 + 3(-3) ........

       4, 4 - 3, 4 - 6, 4 - 9 .........

       4, 1, -2, - 5 ............

 Hence required A.P is  4, 1, -2, - 5 ............ 

(ii)  general form of an A.P is 

      a, a + d, a + 2d, a + 3d ..............

      7, 7 + 13, 7 + 2(13), 7 + 3(13) .........

      7, 20, 33, 46 ................ 

   Hence required A.P is  7, 20, 33, 46 ................ 

Finding First Term And Common difference:

Question1: For the following APs, write the first term and the common difference:

(i) 3, 1, – 1, – 3, . . .                   (ii) – 5, – 1, 3, 7, . . .

Solution: 

(i)  a = 3, 

     d = a_n - a_{n -1} 

  ∴ d = a₂ - a₁

          = 1 - 3 = - 2 

    First term = 3, common difference = - 2

(ii) – 5, – 1, 3, 7, . . .

   a = - 5

   d = a₂ - a₁

      = - 1 - (-5)

      = - 1 + 5

      = 4

 First term = - 5, common difference = 4

Finding n^th term And Number of term:

Question1: From the given APs, find the n^th term.

(i)  17, 13, 9, 5 ........................  33^th term

(ii)  - 5, 0, 5, 10 ...................  24^th term

Solution: 

(i) 17, 13, 9, 5 ........................ 

    a = 17, d = 13 - 17 = - 4, n = 33

a_n = a + (n -1)d

a₃₃ = a + (33 - 1)d

      = a + 32d

      = 17 + 32(-4) 

      = 17 + (-128)

      = 17 - 128

      = - 111

 ∴ 33^th term is - 111.

(ii) - 5, 0, 5, 10 ...................  24^th term

    a = -5,

    d = 0 -(-5) = 0 +5 = 5, 

     n = 24,

  a₂₄ = a + 23d

       = - 5 + 23 (5)

      = - 5 + 115

     = 110

Question2: Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

Solution: 

Method 1: Solving from first term.

    a = 3, d = 8 - 3 = 5, a_n = 253

  // Here we have to find first n of final term 253 //

  a_n = a + (n -1)d

253 = 3 + (n - 1)5

253 - 3  = (n -1)5 

250 = (n - 1)5 

n - 1 = 250 / 5 

n - 1 = 50 

  n = 50 + 1 

  n = 51 

We have to find 20^th term from last term

∴ required n will be 51 - 19 = 32 

a₃₂ = a + 31d 

      = 3 + 31(5) 

     = 3 + 155 = 158

Hence 20^th term from last term is 158

Method 2: solving from final term.

   a = 253, d = 3 - 8 = - 5, n = 20

   a₂₀ = a + 19d 

         = 253 + 19(-5)

         = 253 - 95

         = 158

Hence 20^th ​term from last term is 158