Arithmetic Progressions

Introduction

We have observed many things in nature that follow a certain pattern, such as the holes of a honeycomb, the grains on a maize cob, the spirals on a pineapple etc.

We often observe interesting patterns of numbers in which quantity increases or decreases progressively in our day-to-day life. Some such examples are

Example 1: Ria applied for a job and got selected. She has been offered a job with a starting monthly salary of rupees 10,000 with an annual increment of rupees 1000 in her salary. Her salary (in rupees) for the 1st, 2nd, 3^rd …… years will be respectively.

10,000, 11,000, 12,000, and so on

Example 2: In a saving scheme the amount becomes 2 times of itself after every 3 years. The maturity amount (in rupees) of an investment of rupees 10,000 after 3, 6, 9 and 12 years will be, respectively: 20,000, 40,000, 80,000, and so on

In the examples above, we observed that the succeeding terms are obtained by adding a fixed number in the first example and by multiplying with a fixed number in the second example. These patterns are generally known as sequences.

Therefore, a sequence is an arrangement of numbers in a definite order according to some rule. Example: 2, 4, 6, 8, 10… Here numbers are even natural number.

Now we will study a particular type of sequences which are known as arithmetic progressions.

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.

Example 1: -3, -2, -1, 0 …..

Each of the numbers in the list is called a term. We denote the terms of a sequence by a₁, a₂, a₃ ….etc. or x₁, x₂, x₃ …etc. Here, the subscripts denote the positions of the terms. The first term of the A.P is denoted by a₁ and the number at the second place is called the second term and is denoted by a₂ and so on. In general, the number at the nth place is called the nth term of the sequence and is denoted by a_n. The term is called the general term of the sequence. In the above example a₁ (first term) = -3, a₂ (second term) = -2, a₃ (third term) = -1, a₄ (fourth term) = 0, and so on.

Each term is obtained by adding a fixed number except first term. Here in the above example each term is obtained by adding 1 to the term preceding it. And we can write the other next term by adding same fixed number. This fixed number is called the common difference of the A.P (Arithmetic Progression). It can be positive, negative or zero. Remember that we can find d using any two consecutive terms, once we know that the numbers are in AP. In the above example d = a_n –a_n-1 (difference between the next term and previous term of the A.P.) i.e. a2 – a1 = -2 – (-3) = 1

Example 2: Let us consider the sequence of even natural numbers 2, 4, 6, 8, 10 …..

Here we have, 

a₁ (first term) = 2 = 2x1

a₂ (second term) = 4 = 2x2

a₃ (third term) = 6 = 2x3

a₄ (fourth term) = 8 = 2x4

a₅ (fifth term) = 10 = 2x5 and so on.

It is evident from this that a_n (nth term) = 2xn = 2n (where n = 1, 2, 3…..)

Example 3: Let us now consider the sequence of squares of natural numbers i.e., 1, 4, 9, 16, 25 ……

Here we have, 

a₁ (first term) = 1 = 1²

a₂ (second term) = 4 = 2²

a₃ (third term) = 9 = 3²

a₄ (fourth term) = 16 = 4²

a₅ (fifth term) = 25 = 5² and so on.

It is evident from this that a_n (nth term) = n² (where n = 1, 2, 3…..)

Example 4: Let us study the sequence of odd natural numbers i.e., 1, 3, 5, 7, 9 ….

Here we have, 

a₁ (first term) = 1 = 2x1-1

a₂ (second term) = 3 = 2x2-1

a₃ (third term) = 5 = 2x3-1

a₄ (fourth term) = 7 = 2x4-1

a₅ (fifth term) = 9 = 2x5-1 and so on.

It is evident from this that a_n (nth term) = 2xn-1 = 2n-1 (where n = 1, 2, 3…..)

We can also describe a sequence by writing the algebraic formula for its nth term or general term. In some cases, the terms of the sequence do not follow some fixed pattern but they are generated by some recursive relation.

Example 5: Consider for instance, the sequence, 1, 2, 3, 5, 8 …..

Here we have, 

a₁ (first term) = 1

a₂ (second term) = 2

a₃ (third term) = 2 = 1+2 = a₁ + a₂

a₄ (fourth term) = 3 = 1+3 = a₂ + a₃

a₅ (fifth term) = 5 = 2+3 = a₃ +a₄ and so on.

It is evident from this that a_n (nth term) = a_n-1 + a_n-2 for n>2

Example 6: For the AP: 3, 1, -1, -3…. write the first term a₁ and the common difference d.

Here, first term (a₁) = 3

Common difference (d) = a_n –a_n-1

= a₂ – a₁

= 1-3

= -2

Example 7: Is 4, 10, 16, 22 ….an A.P?  If they form an A.P, write the next two terms.

Here, 

d = a₂ – a₁ = 10 – 4 = 6

d = a₃ – a₂ = 16 – 10 = 6

d = a₄ – a₃ = 22 – 16 = 6

We can see that all the value of common difference is fixed or equal.

Hence, 4, 10, 16, 22….an A.P and next two terms are 22 + 6 = 28 and 28 + 6 = 32.