Chapter – 5

Arithmetic Progressions

In this post we have given the detailed notes of class 10 Maths Chapter 5 (Arithmetic Progressions) in English. These notes are useful for the students who are going to appear in class 10 board exams.

Chapter 5: Arithmetic Progressions

What is an Arithmetic Progression?

In mathematics, an Arithmetic Progression (A.P.) is a list of numbers in which each term, after the first, is obtained by adding a fixed number to the preceding term. This fixed number is called the common difference of the A.P.

In other words, a sequence a₁, a₂, a₃, a₄, … aₙ is called an arithmetic progression if the difference between any two consecutive terms is the same (or constant). Each number in the list is called a term. Here a₁ is the first term, and aₙ is the n^th term of the progression.

Common Difference

The fixed number that is added to (or subtracted from) each term to get the next term is called the common difference, denoted by d. The common difference of an A.P. can be positive, negative, or zero.

If the first term of an A.P. is a₁, the second term is a₂, … and the n^th term is aₙ, with common difference d, then:

a₂ − a₁ = a₃ − a₂ = … = aₙ − aₙ₋₁ = d

This means an A.P. can be generated by repeatedly adding d to the first term:

a, a + d, a + 2d, a + 3d, …

General Form of an A.P.

a, a + d, a + 2d, a + 3d, … represents an arithmetic progression, where a is the first term and d is the common difference. This is called the general form of an A.P.

Worked Examples

If the first term a = 6 and the common difference d = 3, then 6, 9, 12, 15, … is an A.P.

If a = 6 and d = −3, then 6, 3, 0, −3, … is an A.P.

Worked Example — Taxi Fare

Question: The taxi fare after each kilometre, when the fare for the first kilometre is ₹15 and the fare for each additional kilometre is ₹8.

Solution: The fare for the first kilometre, ₹15, is the first term a₁ of the A.P. The fare of ₹8 for each additional kilometre is the common difference d.

The A.P. is a₁, a₁ + d, a₁ + 2d, a₁ + 3d, …
= 15, 15 + 8, 15 + 2 × 8, 15 + 3 × 8, …
= 15, 23, 31, 39, …

This is an arithmetic progression.

Worked Example — Finding the Next Terms

Question: Write the next two terms of the A.P. 4, 10, 16, 22, …

Solution: a₂ − a₁ = 10 − 4 = 6, and a₃ − a₂ = 16 − 10 = 6. So d = 6.

Hence the next two terms are 22 + 6 = 28 and 28 + 6 = 34.

Worked Example — Well-Digging Cost

Question: The cost of digging a well, after every metre of digging, when the cost of digging the first metre is ₹150, and it increases by ₹50 for each subsequent metre.

Solution: Cost of digging the first metre = ₹150, so a₁ = 150. The cost increases by ₹50 for each additional metre, so d = 50.

The A.P. is: 150, 150 + 50, 150 + 2 × 50, 150 + 3 × 50, …
= 150, 200, 250, 300, …

This is an arithmetic progression.

Types of Arithmetic Progressions

Arithmetic progressions are mainly of two types:

1. Finite A.P.

An A.P. that has a limited (fixed) number of terms is called a finite A.P. This type of A.P. has a last term.

Example: 5, 10, 15, 20, 25, 30, … , 100 (last term).

2. Infinite A.P.

An A.P. that has an infinite number of terms is called an infinite A.P. This type of A.P. does not have a last term.

Example: 10, 20, 30, 40, 50, 60, … is an infinite A.P.

n^th Term of an A.P. (General Term)

We know the general form of an A.P. is written as:

a, a + d, a + 2d, a + 3d, a + 4d, …, a + (n − 1)d

Here the first term is a. To find the second term, we add the common difference d to the first term a — or we can say we multiply d by (2 − 1) and add it to a:

a₂ = a + d = a + (2 − 1)d

To find the third term, similarly, we multiply d by (3 − 1) and add it to a:

a₃ = a + 2d = a + (3 − 1)d

In the same way, to find the n^th term of an A.P., we multiply the common difference d by (n − 1) and add it to the first term a, as shown in the general form:

aₙ = a + (n − 1)d

Here, aₙ is called the n^th term, also known as the general term of the A.P.

If an A.P. has m terms, then aₘ is its last term, which is sometimes also denoted by l.

Worked Example — Finding a Term

Question: Find the 10^th term of the A.P. 2, 7, 12, …

Solution: Here a₁ = 2, a₂ = 7, so d = a₂ − a₁ = 7 − 2 = 5.

Since aₙ = a + (n − 1)d,
a₁₀ = a₁ + (10 − 1)d = 2 + 9 × 5 = 47.

Hence the 10^th term is 47.

Worked Example — Finding Which Term

Question: Which term of the A.P. 21, 18, 15, … is −81? Also, is any term of this A.P. equal to zero? Give a reason for your answer.

Solution: Here a = 21, d = 18 − 21 = −3, and aₙ = −81. We need to find n.

Since aₙ = a + (n − 1)d,
−81 = 21 + (n − 1)(−3)
⇒ −81 = 24 − 3n
⇒ −105 = −3n
⇒ n = 35

So the 35^th term of the given A.P. is −81.

Now, to check if any term equals zero, suppose aₙ = 0:
21 + (n − 1)(−3) = 0
⇒ 3(n − 1) = 21
⇒ n = 8

So the 8^th term of this A.P. is 0.

Worked Example — Determining an A.P.

Question: Determine the A.P. whose 3^rd term is 5 and 7^th term is 9.

Solution: We have:
a₃ = a + 2d = 5 … (1)
a₇ = a + 6d = 9 … (2)

Solving equations (1) and (2), we get a = 3, d = 1.

Hence the required A.P. is: 3, 4, 5, 6, 7, …

Sum of the First n Terms of an A.P.

We can derive a formula for the sum of the first n terms of an A.P.

Let an A.P. with first term a and common difference d have n terms written as:

a, a + d, a + 2d, …, a + (n − 1)d

Let the sum of the first n terms be denoted by Sₙ:

Sₙ = a + (a + d) + (a + 2d) + … + [a + (n − 2)d] + [a + (n − 1)d] … (1)

Writing the same terms in reverse order:

Sₙ = [a + (n − 1)d] + [a + (n − 2)d] + … + (a + 2d) + (a + d) + a … (2)

Adding equations (1) and (2), each pair of corresponding terms adds up to [2a + (n − 1)d], and there are n such pairs:

2Sₙ = n × [2a + (n − 1)d]

Sₙ = (n/2)[2a + (n − 1)d]

This is the formula for the sum of the first n terms of an A.P.

Since aₙ = a + (n − 1)d, the formula can also be written as:

Sₙ = (n/2)[a + aₙ] = (n/2)[a + l]

Note: This second form is useful when the first and last terms of the A.P. are known, but the common difference is not given.

Relation Between nth Term and Sum

The n^th term of an A.P. equals the difference between the sum of its first n terms and the sum of its first (n − 1) terms:

aₙ = Sₙ − Sₙ₋₁

Worked Example — Sum of Terms

Question: Find the sum of the first 22 terms of the A.P. 8, 3, −2, …

Solution: Here a = 8, d = 3 − 8 = −5, n = 22.

We know Sₙ = (n/2)[2a + (n − 1)d]

S₂₂ = 22/2 [2 × 8 + (22 − 1)(−5)] = 11(16 − 105) = 11(−89) = −979

So the sum of the first 22 terms is −979.

Worked Example — Finding a Term from the Sum

Question: If the sum of the first 14 terms of an A.P. is 1050 and its first term is 10, find the 20^th term.

Solution: Here S₁₄ = 1050, n = 14, a = 10.

Since Sₙ = (n/2)[2a + (n − 1)d]:
1050 = 14/2 [20 + 13d]
⇒ 910 = 91d
⇒ d = 10

a₂₀ = 10 + (20 − 1) × 10 = 200. So the 20^th term is 200.

Worked Example — Number of Terms

Question: How many terms of the A.P. 24, 21, 18, … must be taken so that their sum is 78?

Solution: Here a = 24, d = 21 − 24 = −3, and Sₙ = 78. We need to find n.

Since Sₙ = (n/2)[2a + (n − 1)d]:
78 = n/2 [48 + (n − 1)(−3)] = n/2 [51 − 3n]
⇒ 3n² − 51n + 156 = 0
⇒ n² − 17n + 52 = 0
⇒ (n − 4)(n − 13) = 0

So n = 4 or n = 13. Both values are possible and acceptable. So the required number of terms is either 4 or 13.

Sum of the First n Positive Integers

The formula for the sum of the first n positive integers can also be derived from this: Let Sₙ = 1 + 2 + 3 + … + n. Here a = 1 and l = n.

Sₙ = n(1 + n)/2 = n(n + 1)/2

Applications of Arithmetic Progressions

Arithmetic progressions are used to generalise a set of patterns that we observe in our daily life — for example, in preparing food, calculating distance, time and cost for travel, understanding loans for cars, trucks, houses or education, and understanding sports statistics.

Sequences and series play an important role in various aspects of our life. They help us predict, evaluate, and monitor the outcome of a situation or event, and are very useful in decision-making.

Worked Example — Sum and Product of Terms

Question: The sum of the 3^rd and 7^th terms of an A.P. is 6, and their product is 8. Find the sum of the first 16 terms of the A.P.

Solution: Let the first term of the A.P. be a₁ and the common difference be d.

a₃ = a₁ + 2d, a₇ = a₁ + 6d

According to the question:
a₃ + a₇ = a₁ + 2d + a₁ + 6d = 6
⇒ a₁ + 4d = 3 … (1)

Also, a₃ × a₇ = (a₁ + 2d)(a₁ + 6d) = 8
⇒ a₁² + 8a₁d + 12d² = 8 … (2)

From (1), a₁ = 3 − 4d. Substituting in (2):
(3 − 4d)² + 8(3 − 4d)d + 12d² = 8
⇒ d = ½ or d = −½

Substituting back in (1): a₁ = 1 or a₁ = 5.

For a₁ = 1, d = ½: S₁₆ = 8[2 + 15 × ½] = 4[4 + 15] = 76

For a₁ = 5, d = −½: S₁₆ = 8[10 + 15 × (−½)] = 4[20 − 15] = 20

So there are two possible values of S₁₆: 76 and 20, corresponding to the two different possible values of a₁ and d.

Worked Example — Cash Prizes

Question: A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.

Solution: Here n = 7, d = −20, S₇ = 700. Let the first prize be ₹a.

S₇ = 7/2[2a + (7 − 1)(−20)] = 700
⇒ 2a − 120 = 200
⇒ a = 320/2 = 160

So the first prize is ₹160. The other prizes are ₹140, ₹120, ₹100, ₹80, ₹60, and ₹40 respectively.

Worked Example — Given d and Last Term

Question: Find the sum of the first 22 terms of an A.P. in which d = 7 and the 22^nd term is 149.

Solution: Here d = 7 and a₂₂ = 149.

Since aₙ = a₁ + (n − 1)d:
a₂₂ = a₁ + (22 − 1)(7) = 149
⇒ a₁ = 149 − 147 = 2

Now, S₂₂ = 22/2 [4 + 21 × 7] = 11 × 151 = 1661.

Hence the sum of the first 22 terms is 1661.

Key Points to Remember

• An A.P. is a list of numbers in which every term (except the first) is obtained by adding a fixed number d to the term preceding it. This fixed number d is called the common difference.
• The general form of an A.P. is a, a + d, a + 2d, a + 3d, …
• A given list of numbers is an A.P. if the differences a₂ − a₁, a₃ − a₂, a₄ − a₃, … are all equal, i.e. aₖ − aₖ₋₁ is the same for all values of k.
• The n^th term (or general term) of an A.P. with first term a and common difference d is: aₙ = a + (n − 1)d.
• Sum of the first n terms of an A.P.: Sₙ = (n/2)[2a + (n − 1)d].
• Sum can also be found using Sₙ = (n/2)[a + l], where l is the last term.
• Relation between n^th term and sum: aₙ = Sₙ − Sₙ₋₁.
• Sum of the first n positive integers: Sₙ = n(n + 1)/2.
• The common difference d can be positive, negative, or zero.

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