Arithmetic Progression

Arithmetic Progression : Formulas, Theorems & Examples (Class 10 Maths)

Sequences, Series, and Progressions: Building Blocks of AP

Before we understand Arithmetic Progression, let’s clarify some fundamental terms:

Sequences:

Definition: A sequence is an ordered list of numbers. Each number in the sequence is called a term. Sequences follow a specific rule or pattern to determine the next term.

Types:

  • Finite Sequence: A sequence with a limited number of terms. It has a last term.

Example: 2, 4, 6, 8, 10 (This is a finite sequence with 5 terms)

  • Infinite Sequence: A sequence with an unlimited number of terms. It continues indefinitely and does not have a last term.

Example: 1, 2, 3, 4, 5, … (The “…” indicates that the sequence continues infinitely)

Examples of Sequences:

  • Sequence of even numbers: 2, 4, 6, 8, 10, …
  • Sequence of odd numbers: 1, 3, 5, 7, 9, …
  • Sequence of squares of natural numbers: 1, 4, 9, 16, 25, …
  • Sequence defined by a rule: Let the nth term be given by an = 2n + 1. Then the sequence is: 3, 5, 7, 9, 11, … (by substituting n=1, 2, 3, 4, 5, …)

Series:

Definition: A series is the sum of the terms of a corresponding sequence.

Formation: If a sequence is a1, a2, a3, … , an, …, then the series is a1 + a2 + a3 + … + an + …

Types (corresponding to sequences):

  • Finite Series: Sum of terms of a finite sequence.

Example: For the sequence 2, 4, 6, 8, 10, the series is 2 + 4 + 6 + 8 + 10 = 30

  • Infinite Series: Sum of terms of an infinite sequence.

Example: For the sequence 1, 2, 3, 4, 5, …, the series is 1 + 2 + 3 + 4 + 5 + …

Examples of Series:

  • Series of first five even numbers: 2 + 4 + 6 + 8 + 10
  • Series of natural numbers: 1 + 2 + 3 + 4 + 5 + …

Progressions:

Definition: A progression is a special type of sequence where the terms follow a specific mathematical pattern that can be expressed using a formula for the general term (or nth term). Essentially, there’s a predictable rule to generate the terms.

Examples of Progressions:

  • Arithmetic Progression (AP): The difference between consecutive terms is constant (what we’ll focus on).
  • Geometric Progression (GP): The ratio between consecutive terms is constant.
  • Harmonic Progression (HP): The reciprocals of the terms are in Arithmetic Progression.

Arithmetic Progression (AP)

Definition of Arithmetic Progression (AP):

An Arithmetic Progression (AP) is a progression (and therefore a sequence) in which the difference between any two consecutive terms is constant. This constant difference is the defining characteristic of an AP.

Common Difference (d): This constant difference is called the common difference, denoted by ‘d’.

Understanding Common Difference:

If a1, a2, a3, a4, … is an AP, then:

a2 – a1 = d a3 – a2 = d a4 – a3 = d and so on…

In general, a(n+1) – an = d for any term an and its succeeding term a(n+1).

Example of Arithmetic Progression:

Consider the sequence: 2, 5, 8, 11, 14, …

Let’s check the difference between consecutive terms:

5 – 2 = 3 8 – 5 = 3 11 – 8 = 3 14 – 11 = 3

The difference is consistently 3. Therefore, 2, 5, 8, 11, 14, … is an Arithmetic Progression with a common difference (d) = 3.

More Examples of Arithmetic Progressions:

  • Example 1: 10, 8, 6, 4, 2, 0, -2, …

Common difference: 8 – 10 = -2, 6 – 8 = -2, and so on. Here, d = -2.

  • Example 2: -3, -3, -3, -3, -3, …

Common difference: -3 – (-3) = 0, -3 – (-3) = 0, and so on. Here, d = 0.

  • Example 3: 0.5, 1.0, 1.5, 2.0, 2.5, …

Common difference: 1.0 – 0.5 = 0.5, 1.5 – 1.0 = 0.5, and so on. Here, d = 0.5.

 

Impact of Common Difference (d) on AP:

  • Positive Common Difference (d > 0): The AP is increasing. Each term is greater than the previous term.

Example: 2, 5, 8, 11, … (d = 3)

  • Zero Common Difference (d = 0): The AP is constant. All terms are the same.

Example: -3, -3, -3, -3, … (d = 0)

  • Negative Common Difference (d < 0): The AP is decreasing. Each term is smaller than the previous term.

Example: 10, 8, 6, 4, … (d = -2)

Finite and Infinite Arithmetic Progressions: Bounded and Unbounded

Finite AP:

Definition: A finite Arithmetic Progression is an AP that has a finite (countable) number of terms. It has a last term.

Example: The AP: 2, 5, 8, …, 32, 35, 38

Here, the first term is 2, the common difference is 3, and the last term is 38. We can count the number of terms (though it might be tedious).

Infinite AP:

Definition: An infinite Arithmetic Progression is an AP that has an infinite (uncountable) number of terms. It continues indefinitely and does not have a last term.

Example: The AP: 2, 5, 8, 11, …

Here, the first term is 2, the common difference is 3, and the “…” indicates that it goes on forever.

Key Difference:

  • Finite AP: Has a last term, number of terms is countable.
  • Infinite AP: No last term, number of terms is uncountable (infinite).

General Term (Nth Term) of an AP: The Formula to Find Any Term

Sometimes, we need to find a specific term in an AP, like the 10th term, 50th term, or even the 1000th term. Listing out all terms to reach that position would be very time-consuming.

The general term formula provides a direct way to calculate any term in an AP without listing all preceding terms.

Formula for the Nth Term (Tn or an):

Let:

  • a be the first term of the AP.
  • d be the common difference.
  • n be the position of the term we want to find (e.g., n=1 for the first term, n=2 for the second term, etc.).
  • Tn (or an) represent the nth term of the AP.

Then, the formula for the nth term is:

T_n = a + (n – 1)d

Explanation of the Formula:

  • For the 1st term (n=1): T1 = a + (1 – 1)d = a + 0d = a (which is correct, the first term is ‘a’)
  • For the 2nd term (n=2): T2 = a + (2 – 1)d = a + 1d = a + d (which is also correct, the second term is the first term plus the common difference)
  • For the 3rd term (n=3): T3 = a + (3 – 1)d = a + 2d = a + d + d (and so on)

Examples using the Nth Term Formula:

  • Example 1: Find the 10th term of the AP: 2, 5, 8, …

Here, a = 2, d = 3, and we want to find the 10th term, so n = 10.

T_10 = a + (10 – 1)d = 2 + (9) * 3 = 2 + 27 = 29

Therefore, the 10th term is 29.

  • Example 2: Which term of the AP: 21, 18, 15, … is -81? Also, is 0 any term of this AP?

Here, a = 21, d = 18 – 21 = -3, and we are given Tn = -81. We need to find ‘n’.

-81 = 21 + (n – 1)(-3) -81 – 21 = (n – 1)(-3) -102 = (n – 1)(-3) -102 / -3 = n – 1 34 = n – 1 n = 34 + 1 = 35

So, the 35th term is -81.

Now, let’s check if 0 is any term. Let Tn = 0.

0 = 21 + (n – 1)(-3) -21 = (n – 1)(-3) -21 / -3 = n – 1 7 = n – 1 n = 7 + 1 = 8

Since ‘n’ is a positive integer (8), 0 is the 8th term of this AP.

 

Sum of Terms in an AP:

Sometimes, we need to find the sum of the first ‘n’ terms of an AP. There are two common formulas for this:

Formula 1 (using first term ‘a’, common difference ‘d’, and number of terms ‘n’):

S_n = n/2 * [2a + (n – 1)d]

Formula 2 (using first term ‘a’, last term ‘l’, and number of terms ‘n’):

If we know the last term (let’s denote it as ‘l’ or Tn), then a simpler formula is:

S_n = n/2 * (a + l)

Where ‘l’ (last term) = Tn = a + (n – 1)d. Formula 2 is derived from Formula 1 by substituting l = a + (n-1)d.

Examples using Sum Formulas:

  • Example 1: Find the sum of the first 20 terms of the AP: 2, 5, 8, …

Here, a = 2, d = 3, n = 20.

Using Formula 1:

S_20 = 20/2 * [2(2) + (20 – 1)3] S_20 = 10 * [4 + (19)3] S_20 = 10 * [4 + 57] S_20 = 10 * 61 = 610

The sum of the first 20 terms is 610.

  • Example 2: Find the sum of the AP: 7, 10.5, 14, …, 84.

Here, a = 7, d = 10.5 – 7 = 3.5, and the last term l = 84. We first need to find ‘n’ (number of terms).

l = a + (n – 1)d 84 = 7 + (n – 1)3.5 84 – 7 = (n – 1)3.5 77 = (n – 1)3.5 77 / 3.5 = n – 1 22 = n – 1 n = 23

Now, using Formula 2:

S_23 = 23/2 * (a + l) S_23 = 23/2 * (7 + 84) S_23 = 23/2 * (91) S_23 = (23 * 91) / 2 = 2093 / 2 = 1046.5

The sum of the AP is 1046.5.

 

Arithmetic Mean (A.M): The Average Value

The Arithmetic Mean (A.M.) of a given set of numbers is simply their average. It is calculated by dividing the sum of the numbers by the total count of numbers.

Formula:

For a set of numbers x1, x2, x3, …, xn, the Arithmetic Mean (A.M.) is:
A.M. = (Sum of terms) / (Number of terms) = (x1 + x2 + x3 + … + xn) / n

Arithmetic Mean in AP:

While the Arithmetic Mean is defined for any set of numbers (they don’t need to be in AP), it has a special property in the context of APs.

For any three consecutive terms in an AP, say a, b, c, the middle term ‘b’ is the Arithmetic Mean of ‘a’ and ‘c’.

b = (a + c) / 2

This is because in an AP, b – a = c – b = d (common difference). From b – a = c – b, we get 2b = a + c, or b = (a + c) / 2.

Example of Arithmetic Mean:

Find the Arithmetic Mean of the numbers: 5, 10, 15, 20, 25.

Sum of terms = 5 + 10 + 15 + 20 + 25 = 75 Number of terms = 5 A.M. = 75 / 5 = 15

The Arithmetic Mean is 15. Notice that in this AP (5, 10, 15, 20, 25), the middle term is indeed 15, which is the A.M.

 

Basic Adding Patterns in an AP: Equidistant Terms

In an AP, the sum of two terms that are equidistant from the beginning and the end of the AP is always constant and equal to the sum of the first and last terms.

Algebraically: For a finite AP with ‘n’ terms, if T1, T2, T3, …, Tn are the terms, then:

T_1 + T_n = T_2 + T_(n-1) = T_3 + T_(n-2) = … = constant

In general, Tr + T(n-r+1) = constant for any r from 1 to n.

Example Illustrating Equidistant Terms Property:

Consider the AP: 2, 5, 8, 11, 14, 17, 20 (Here, n = 7 terms)

T_1 = 2, T_7 = 20 => T_1 + T_7 = 2 + 20 = 22 T_2 = 5, T_6 = 17 => T_2 + T_6 = 5 + 17 = 22 T_3 = 8, T_5 = 14 => T_3 + T_5 = 8 + 14 = 22 T_4 = 11, T_4 (middle term) – we can consider T_4 + T_4 = 2T_4 = 2*11 = 22

The sum is consistently 22.

Sum of First n Natural Numbers:

The sequence of natural numbers: 1, 2, 3, 4, 5, …, n, … is an Arithmetic Progression.

  • First term (a) = 1
  • Common difference (d) = 2 – 1 = 1

Formula for the Sum of First n Natural Numbers:

The sum of the first ‘n’ natural numbers is given by a specific formula, which is derived by treating the natural numbers as an AP:

S_n = n(n + 1) / 2

Derivation (using AP sum formula):

Using the AP sum formula: Sn = n/2 * [2a + (n – 1)d]

For natural numbers, a = 1 and d = 1. Substituting these values:

S_n = n/2 * [2(1) + (n – 1)1] S_n = n/2 * [2 + n – 1] S_n = n/2 * [n + 1] S_n = n(n + 1) / 2

Example using Sum of First n Natural Numbers Formula:

Find the sum of the first 100 natural numbers (1 + 2 + 3 + … + 100).

S_100 = 100(100 + 1) / 2 = 100 * 101 / 2 = 50 * 101 = 5050

The sum is 5050.

Summary of Key Takeaways:

  • Arithmetic Progression (AP) is a sequence with a constant difference between consecutive terms (common difference ‘d’).
  • Types of APs: Finite (has last term) and Infinite (no last term).
  • Nth term formula: Tn = a + (n – 1)d.
  • Sum of first n terms formulas: Sn = n/2 * [2a + (n – 1)d] or Sn = n/2 * (a + l).
  • Arithmetic Mean (A.M.) is the average of numbers. For AP terms a, b, c, b = (a + c) / 2.
  • Sum of equidistant terms from ends of an AP is constant.
  • Sum of first n natural numbers: Sn = n(n + 1) / 2.