NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions are designed to help students understand every topic of the chapter in a clear and simple way. Arithmetic Progressions is an important chapter in the Class 10 Maths syllabus and is often asked in school exams and board examinations. In this chapter, students learn about number patterns in which the difference between two consecutive terms remains the same. Such a sequence is called an arithmetic progression, or A.P.

These Class 10 Maths Chapter 5 solutions explain all the important concepts with easy steps and correct formulas. Students can learn how to identify whether a given sequence is an arithmetic progression, find the first term, calculate the common difference and write the next terms of an A.P. The solutions also explain how to use the formula for the nth term of an arithmetic progression. This formula helps students find any required term without writing the complete sequence.

The chapter also teaches students how to calculate the sum of the first n terms of an arithmetic progression. The NCERT solutions provide step-by-step methods for using the sum formulas correctly. They also include answers to word problems based on daily-life situations, such as savings, salaries, arrangements, distances and number patterns. Each solution is written in a simple manner so that students can understand the method and apply it to similar questions.

The exercise-wise NCERT Solutions for Arithmetic Progressions cover Exercises 5.1, 5.2, 5.3 and 5.4. Students can use these answers to complete homework, check their solutions, clear doubts and prepare for tests. The detailed working also helps students understand where each value comes from and why a particular formula is used.

These solutions are useful for quick revision before exams. Students can review important formulas, practise different types of questions and improve their calculation skills. Regular practice from the NCERT textbook can build confidence and help students score better marks in Maths.

Download Class 10 Maths Chapter 5 Arithmetic Progressions NCERT Solutions PDF

Download the Class 10 Maths Chapter 5 Arithmetic Progressions NCERT Solutions PDF to access clear, step-by-step answers to all the questions from the chapter. These solutions cover every exercise and explain important concepts such as the common difference, nth term, sum of an arithmetic progression, and related word problems in an easy-to-understand manner. The PDF is useful for homework, exam preparation, revision, and strengthening your understanding of Arithmetic Progressions.

Access NCERT Solutions for Maths Chapter 5 – Arithmetic Progression

Exercise 5.1

Q1. Decide whether the numerical pattern in each situation is an arithmetic progression, and explain your reasoning.

(i). The Taxi Fare After Each Km When the Fare is ₹15 for the First Km and ₹8 for Each Additional Km.

Solution: Begin by writing the known quantities. The fare of first km is ₹15 and the fare for each additional km is ₹8. Hence,

Taxi fare for 1st km is ₹15.

Taxi fare for 2nd km is ₹15 + 8 = 23.

Taxi fare for 3rd km is ₹23 + 8 = 31.

Following the same pattern, Taxi fare for nth km is ₹15 + (n − 1) 8.

Thus, this sequence forms an A.P with common difference of 8.

(ii). The Amount of Air Present in a Cylinder When a Vacuum Pump Removes a Quarter of the Air Remaining in the Cylinder at a Time.

Solution: Let the initial volume of air in a cylinder be V liter. In each stroke, the vacuum pump removes 1 4 of air remaining in the cylinder at a time. Hence,

Volume after 1st stroke is 3 V 4.

Volume after 2nd stroke is 3 4 (3 V 4).

Volume after 3rd stroke is (3 4) 2 (3 V 4).

Following the same pattern, Volume after nth stroke is (3 4) n V.

Notice that the subsequent terms are not added with a constant digit but are being multiplied by 3 4. Thus, this sequence does not form an A.P.

(iii). The cost of digging a well after every meter of digging, when it costs ₹150 for the first meter and rises by ₹50 for each subsequent meter.

Solution: First identify the relevant values. The cost of digging for the first meter is ₹150 and the cost for each additional meter is ₹50. Hence,

Cost of digging for 1st meter is ₹150.

Cost of digging for 2nd meter is ₹150 + 50 = 200.

Cost of digging for 3rd meter is ₹200 + 50 = 250.

Following the same pattern, Cost of digging for nth meter is ₹150 + (n − 1) 50.

Thus, this sequence forms an A.P with common difference of 50.

(iv). The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8 % per annum.

Solution: Start with the information given in the question. The principal amount is ₹10000 and the compound interest is 8 % per annum. Hence,

Amount after 1st year is ₹10000 (1 + 8 100).

Amount after 2nd year is ₹10000 (1 + 8 100) 2.

Amount after 3rd year is ₹10000 (1 + 8 100) 3.

Following the same pattern, Amount after nth year is ₹10000 (1 + 8 100) n.

Notice that the subsequent terms are not added with a constant digit but are being multiplied by (1 + 8 100). Thus, this sequence does not form an A.P.

Q2. Write the first four terms of each A.P. using the given first term a and common difference d:

1. a = 10, d = 10

Solution: Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (1)

Substituting a = 10, d = 10 in (1) this gives: a n = 10 + 10 (n − 1) = 10 n (2)

Hence, from (2)

a 1 = 10, a 2 = 20, a 3 = 30 and a 4 = 40.

2. a = − 2, d = 0

Solution: Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (1)

Substituting a = − 2, d = 0 in (1) the result is: a n = − 2 + 0 (n − 1) = − 2 (2)

Hence, from (2)

a 1 = − 2, a 2 = − 2, a 3 = − 2 and a 4 = − 2.

3. a = 4, d = − 3

Solution: Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (1)

Substituting a = 4, d = − 3 in (1) we have: a n = 4 − 3 (n − 1) = 7 − 3 n (2)

Hence, from (2)

a 1 = 4, a 2 = 1, a 3 = − 2 and a 4 = − 5.

4. a = − 1, d = 1 / 2

Solution: Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (1)

Substituting a = − 1, d = 1 / 2 in (1) we obtain: a n = − 1 + 1 2 (n − 1) = n − 3 2 (2)

Hence, from (2)

a 1 = − 1, a 2 = − 1 2, a 3 = 0 and a 4 = 1 2.

5. a = − 1.25, d = − 0.25

Solution: Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (1)

Substituting a = − 1.25, d = − 0.25 in (1) this gives: a n = − 1.25 − 0.25 (n − 1) = − 1 − 0.25 n (2)

Hence, from (2)

a 1 = − 1.25, a 2 = − 1.5, a 3 = − 1.75 and a 4 = − 2.

Q3. For each arithmetic progression, identify the first term and its common difference.

1. 3, 1, − 1, − 3, …

Solution: Reading the sequence, the first term is 3.

The common difference is obtained by subtracting a term from the term immediately after it.

Common difference =

2nd

term − 1st

term

∴ Common difference =

1 − 3 = − 2.

2. − 5, − 1, 3, 7, …

Solution: Reading the sequence, the first term is − 5.

The common difference is obtained by subtracting a term from the term immediately after it.

Common difference =

2nd

term − 1st

term

∴ Common difference =

− 1 − (− 5) = 4.

3. 1 3, 5 3, 9 3, 13 3, …

Solution: Reading the sequence, the first term is 1 3.

The common difference is obtained by subtracting a term from the term immediately after it.

Common difference =

2nd

term − 1st

term

∴ Common difference =

5 3 − 1 3 = 4 3.

4. 0.6, 1.7, 2.8, 3.9, …

Solution: Reading the sequence, the first term is 0.6.

The common difference is obtained by subtracting a term from the term immediately after it.

Common difference =

2nd

term − 1st

term

∴ Common difference =

1.7 − 0.6 = 1.1.

Q4. Check which sequences are arithmetic progressions. For every valid A.P., find d and add the next three terms.

1. 2,4,8,16 …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = 4 − 2 = 2 (1)

a 3 − a 2 = 8 − 4 = 4 (2)

a 4 − a 3 = 16 − 8 = 8 (3)

The calculations above show that the difference between all consecutive terms is not equal.

Hence, the sequence is not an arithmetic progression.

2. 2, 5 2,3, 7 2, …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = 5 2 − 2 = 1 2 (1)

a 3 − a 2 = 3 − 5 2 = 1 2 (2)

a 4 − a 3 = 7 2 − 3 = 1 2 (3)

The calculations above show that the difference between all consecutive terms is equal.

So, the given sequence forms an A.P. with first term 2 and common difference 1 2.

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Substituting a = 2, d = 1 2 in (1) this gives: a n = 2 + 1 2 (n − 1) = n + 3 2 (5)

Hence, from (5)

a 5 = 4, a 6 = 9 2 and a 7 = 5.

3. 1.2, 3.2, 5.2, 7.2 …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = 3.2 − 1.2 = 2 (1)

a 3 − a 2 = 5.2 − 3.2 = 2 (2)

a 4 − a 3 = 7.2 − 5.2 = 2 (3)

The calculations above show that the difference between all consecutive terms is equal.

So, the given sequence forms an A.P. with first term 1.2 and common difference 2.

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Substituting a = 1.2, d = 2 in (1) we obtain: a n = 1.2 + 2 (n − 1) = 2 n − 0.8 (5)

Hence, from (5)

a 5 = 9.2, a 6 = 11.2 and a 7 = 13.2.

4. − 10, − 6, − 2, 2, …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = − 6 − (− 10) = 4 (1)

a 3 − a 2 = − 2 − (− 6) = 4 (2)

a 4 − a 3 = 2 − (− 2) = 4 (3)

The calculations above show that the difference between all consecutive terms is equal.

So, the given sequence forms an A.P. with first term − 10 and common difference 4.

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Substituting a = − 10, d = 4 in (1) we have: a n = − 10 + 4 (n − 1) = 4 n − 14 (5)

Hence, from (5)

a 5 = 6, a 6 = 10 and a 7 = 14.

5. 3, 3 + 2, 3 + 2 2, 3 + 3 2, …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = (3 + 2) − (3) = 2 (1)

a 3 − a 2 = (3 + 2 2) − (3 + 2) = 2 (2)

a 4 − a 3 = (3 + 3 2) − (3 + 2 2) = 2 (3)

The calculations above show that the difference between all consecutive terms is equal.

So, the given sequence forms an A.P. with first term 3 and common difference 2.

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Substituting a = 3, d = 2 in (1) the result is: a n = 3 + (n − 1) 2 (5)

Hence, from (5)

a 5 = 3 + 4 2, a 6 = 3 + 5 2 and a 7 = 3 + 6 2.

6. 0.2,0.22,0.222,0.2222 …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = 0.22 − 0.2 = 0.02 (1)

a 3 − a 2 = 0.222 − 0.22 = 0.002 (2)

a 4 − a 3 = 0.2222 − 0.222 = 0.0002 (3)

The calculations above show that the difference between all consecutive terms is not equal.

Hence, the sequence is not an arithmetic progression.

7. 0, − 4, − 8, − 12…

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = − 4 − 0 = − 4 (1)

a 3 − a 2 = − 8 − (− 4) = − 4 (2)

a 4 − a 3 = − 12 − (− 8) = − 4 (3)

The calculations above show that the difference between all consecutive terms is equal.

So, the given sequence forms an A.P. with first term 0 and common difference − 4.

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Substituting a = 0, d = − 4 in (1) we obtain: a n = 0 − 4 (n − 1) = 4 − 4 n (5)

Hence, from (5)

a 5 = − 16, a 6 = − 20 and a 7 = − 24.

8. − 1 2, − 1 2, − 1 2, − 1 2 …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = (− 1 2) − (− 1 2) = 0 (1)

a 3 − a 2 = (− 1 2) − (− 1 2) = 0 (2)

a 4 − a 3 = (− 1 2) − (− 1 2) = 0 (3)

The calculations above show that the difference between all consecutive terms is equal.

So, the given sequence forms an A.P. with first term − 1 2 and common difference 0.

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Substituting a = − 1 2, d = 0 in (1) we have: a n = − 1 2 + 0 (n − 1) = − 1 2 (5)

Hence, from (5)

a 5 = − 1 2, a 6 = − 1 2 and a 7 = − 1 2.

9. 1, 3, 9, 27, …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = 3 − 1 = 2 (1)

a 3 − a 2 = 9 − 3 = 6 (2)

a 4 − a 3 = 27 − 9 = 18 (3)

The calculations above show that the difference between all consecutive terms is not equal.

10. a, 2 a, 3 a, 4 a, …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = 2 a − a = a (1)

a 3 − a 2 = 3 a − 2 a = a (2)

a 4 − a 3 = 4 a − 3 a = a (3)

The calculations above show that the difference between all consecutive terms is equal.

So, the given sequence forms an A.P. with first term a and common difference a.

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Substituting, a = a, d = a in (4)

this gives: a n = a + (n − 1) d....(5)

a 5 = a + (5 − 1) a = 5 a

a 6 = a + (6 − 1) a = 6 a

a 7 = a + (7 − 1) a = 7 a

11. a, a 2, a 3, a 4, …

Solution: Compare the differences between consecutive terms. a 2 − a 1 = a 2 − a = a (a − 1) (1)

a 3 − a 2 = a 3 − a 2 = a 2 (a − 1) (2)

a 4 − a 3 = a 4 − a 3 = a 3 (a − 1) …(3)

The calculations above show that the difference between all consecutive terms is not equal.

Hence, the sequence is not an arithmetic progression.

12. 2, 8, 18, 32, …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = 8 − 2 = 2 2 − 2 = 2 (1)

a 3 − a 2 = 18 − 8 = 3 2 − 2 2 = 2 …(2)

a 4 − a 3 = 32 − 18 = 4 2 − 3 2 = 2 (3)

The calculations above show that the difference between all consecutive terms is equal.

So, the given sequence forms an A.P. with first term a and common difference is given by a n = a + (n − 1) d (4)

Substituting, a = 2 d = 2 in (4)

we have:

a 5 = 2 + (5 − 1) 2 = 2 + 4 2 = 5 2 = 50

a 6 = 2 + (6 − 1) 2 = 2 + 5 2 = 6 2 = 72

a 7 = 2 + (7 − 1) 2 = 2 + 6 2 = 7 2 = 98

13. 3, 6, 9, 12, …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = 6 − 3 = 3 × 2 − 3 = 3 (2 − 1) (1)

a 3 − a 2 = 9 − 6 = 3 − 6 = 3 (3 − 2) …(2)

a 4 − a 3 = 12 − 9 = 2 3 − 3 = 3 (2 − 3) ….(3)

The calculations above show that the difference between all consecutive terms is not equal.

Hence, the sequence is not an arithmetic progression.

14. 1 2, 3 2, 5 2, 7 2, …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = 3 2 − 1 2 = 8 (1)

a 3 − a 2 = 5 2 − 3 2 = 16 …(2)

a 4 − a 3 = 7 2 − 5 2 = 24 ….(3)

The calculations above show that the difference between all consecutive terms is not equal.

Hence, the sequence is not an arithmetic progression.

15. 1 2, 5 2, 7 2, 73, …

Solution: Compare the differences between consecutive terms.

a 2 − a 1 = 5 2 − 1 2 = 24 (1)

a 3 − a 2 = 7 2 − 5 2 = 24 …(2)

a 4 − a 3 = 73 − 7 2 = 24 ….(3)

The calculations above show that the difference between all consecutive terms is equal.

So, the given sequence forms an A.P. with first term a and common difference is given by a n = a + (n − 1) d (4)

Substituting a = 1 2, d=24 in (4)

this gives:

a 5 = 1 2 + (5 − 1) 24 = 1 + (4) 24 = 97

a 6 = 1 2 + (6 − 1) 24 = 1 + (5) 24 = 121

a 7 = 1 2 + (7 − 1) 24 = 1 + (6) 24 = 145

Exercise 5.2

Q1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a n the nth term of the A.P.

(i). Ans: The first term is a = 7 (1)

The common difference is d = 3 (2)

The term number is n = 8 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Putting the known values into (4) gives

a n = 7 + (8 − 1) 3

⇒ a n = 7 + 21

∴ a n = 28

(ii). Ans: The first term is a = − 18 (1)

The stated nth term is a n = 0 (2)

The term number is n = 10 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Putting the known values into (4) gives

0 = − 18 + (10 − 1) d

⇒ 18 = 9 d

∴ d = 2

(iii). Ans: The stated nth term is a n = − 5 (1)

The common difference is d = − 3 (2)

The term number is n = 18 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Putting the known values into (4) gives

− 5 = a + (18 − 1) (− 3)

⇒ − 5 = a − 51

∴ a n = 46

(iv) Ans: The first term is a = − 18.9 (1)

The common difference is d = 2.5 (2)

The stated nth term is a n = 3.6 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Putting the known values into (4) gives

3.6 = − 18.9 + (n − 1) (2.5)

⇒ 22.5 = (n − 1) (2.5)

⇒ 9 = (n − 1)

∴ a n = 10

(v). Ans: The first term is a = 3.5 (1)

The common difference is d = 0 (2)

The term number is n = 105 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Putting the known values into (4) gives

a n = 3.5 + (105 − 1) (0)

∴ a n = 3.5

Q2. Choose the correct choice in the following and justify

(i). 30th term of the A.P 10, 7, 4, …, is

97

77

77

87

Solution: C. − 77

The first term is a = 10 (1)

The common difference is d = 7 − 10 = − 3 (2)

The term number is n = 30 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Putting the known values into (4) gives a n = 10 + (30 − 1) (− 3)

⇒ a n = 10 − 87

∴ a n = − 77

(ii). 11th term of the A.P − 3, − 1 2, 2, …, is

28

22

38

48 1 2

Solution: B. 22

The first term is a = − 3 (1)

The common difference is d = − 1 2 − (− 3) = 5 2 (2)

The term number is n = 11 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Putting the known values into (4) gives a n = − 3 + 5 2 (11 − 1)

⇒ a n = − 3 + 25

∴ a n = 22

Q3. In the following APs find the missing term in the blanks

(i). 2, _ _, 26

Solution: First identify the relevant values. first term a = 2 (1)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d

Substituting the values from (1) we have: a n = 2 + (n − 1) d (2)

Given, third term a 3 = 26. From (2) we obtain:

26 = 2 + (3 − 1) d

⇒ 26 = 2 + 2 d

∴ d = 12 ….(3)

From (1), (2) and (3) we have for n = 2

a 2 = 2 + (2 − 1) (12)

∴ a 2 = 14

∴ The sequence is 2, 14, 26.

(ii). _ _, 13, _ _, 3

Solution: First identify the relevant values. second term a 2 = 13 (1)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (2)

Substituting the values from (1) for n = 2 we have: 13 = a + d (3)

Given, fourth term a 4 = 3. From (2) we obtain:

3 = a + 3 d (4)

Subtract (3) from (4) and solve:

3 − 13 = (a + 3 d) − (a + d)

⇒ − 10 = 2 d

∴ d = − 5 ….(5)

From (3) and (5) this gives

13 = a − 5

⇒ a = 18 (6)

Substituting the values from (5) and (6) in (2) this gives:

a n = 18 − 5 (n − 1) (7)

First term, a = 18 and third term a 3 = 8

∴ The sequence is 18, 13, 8, 3.

(iii). 5, _ _, _ _, 9 1 2

Solution: Use the stated data to set up the progression. first term a = 5 (1)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (2)

Using (1) in (2) gives a n = 5 + (n − 1) d (3)

Given, fourth term a 4 = 9 1 2. From (3) this gives:

9 1 2 = 5 + (4 − 1) d

⇒ 9 1 2 = 5 + 3 d

∴ d = 3 2 ….(4)

From (3) and (4) this gives

a n = 5 + 3 2 (n − 1) (5)

Second term, a 2 = 13 2 and third term a 3 = 8

∴ The sequence is 5, 13 2, 8, 9 1 2.

(iv). − 4, _ _, _ _, _ _, _ _, 6

Solution: Use the stated data to set up the progression. first term a = − 4 (1)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (2)

Using (1) in (2) gives a n = − 4 + (n − 1) d (3)

Given, sixth term a 6 = 6. From (3) this gives:

6 = − 4 + (6 − 1) d

⇒ 6 = − 4 + 5 d

∴ d = 2 ….(4)

From (3) and (4) this gives

a n = − 4 + 2 (n − 1) (5)

Second term a 2 = − 2, third term a 3 = 0, fourth term a 4 = 2 and fifth term a 5 = 4.

∴ The sequence is − 4, − 2, 0, 2, 4, 6

(v). _ _, 38, _ _, _ _, _ _, − 22

Solution: Use the stated data to set up the progression. second term a 2 = 38 (1)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (2)

Substituting the values from (1) for n = 2 we obtain: 38 = a + d (3)

Given, sixth term a 6 = − 22. From (2) this gives:

− 22 = a + 5 d (4)

Subtract (3) from (4) and solve:

− 22 − 38 = (a + 5 d) − (a + d)

⇒ − 60 = 4 d

∴ d = − 15 ….(5)

From (3) and (5) we obtain

38 = a − 15

⇒ a = 53 (6)

Substituting the values from (5) and (6) in (2) the result is:

a n = 53 − 15 (n − 1) (7)

First term, a = 53, second term a 3 = 23, third term a 3 = 8 and fourth term a 4 = − 7

∴ The sequence is 53, 38, 23, 8, − 7, − 22.

Q4. Which term of the A.P. 3, 8, 13, 18, … is 78?

Solution: Begin by writing the known quantities. the first Term, a = 3 (1)

The common difference is d = 8 − 3 = 5 (2)

The stated nth term is a n = 78 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Putting the known values into (4) gives

78 = 3 + 5 (n − 1)

⇒ 75 = 5 (n − 1)

⇒ 15 = (n − 1)

∴ n = 16

So, 16th term of this A.P. is 78.

Q5. Find the number of terms in each of the following A.P.

(i). 7,13,19, …,205

Solution: Begin by writing the known quantities. the first Term, a = 7 (1)

The common difference is d = 13 − 7 = 6 (2)

The stated nth term is a n = 205 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Putting the known values into (4) gives

205 = 7 + 6 (n − 1)

⇒ 198 = 6 (n − 1)

⇒ 33 = (n − 1)

∴ n = 34

So, given A.P. sequence contains 34 terms.

(ii). 18, 15 1 2, 13, …, − 47

Solution: Use the stated data to set up the progression. the first Term, a = 18 (1)

The common difference is d = 15 1 2 − 18 = − 5 2 (2)

The stated nth term is a n = − 47 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Putting the known values into (4) gives

− 47 = 18 − 5 2 (n − 1)

⇒ − 65 = − 5 2 (n − 1)

⇒ 26 = (n − 1)

∴ n = 27

So, given A.P. sequence contains 27 terms.

Q6. Check whether − 150 is a term of the A.P. 11, 8, 5, 2, …

Solution: First identify the relevant values. the first Term, a = 11 (1)

The common difference is d = 8 − 11 = − 3 (2)

The stated nth term is a n = − 150 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Putting the known values into (4) gives

− 150 = 11 − 3 (n − 1)

⇒ − 161 = − 3 (n − 1)

⇒ 161 3 = (n − 1)

∴ n = 164 3

Since n is not a natural number. So, − 150 is not a term of the given A.P. series.

Q7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.

Solution: Start with the information given in the question. the 11th Term, a 11 = 38 (1)

Given, the 16th Term, a 16 = 73 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Using (1) in (3) gives

38 = a + (11 − 1) d

⇒ 38 = a + 10 d (4)

Using (2) in (3) gives

73 = a + (16 − 1) d

⇒ 73 = a + 15 d (5)

Subtract (4) from (5) to solve (4) and (5):

⇒ 73 − 38 = (a + 15 d) − (a + 10 d)

⇒ 5 = 35 d

∴ d = 7 (6)

Substituting value from (6) in (4) this gives:

⇒ 38 = a + 70

∴ a = − 32 (7)

Substituting once more the values from (6) and (7) in (3) we obtain:

a n = − 32 + 7 (n − 1) (8)

For the 31st term substitute n = 31 in (8) the result is:

a 31 = − 32 + 7 (31 − 1)

⇒ a 31 = − 32 + 210

∴ a 31 = 178

So, the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73 is 178.

Q8. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution: Start with the information given in the question. the 3rd Term, a 3 = 12 (1)

Given, the 50th Term, a 50 = 106 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Using (1) in (3) gives

12 = a + (3 − 1) d

⇒ 12 = a + 2 d (4)

Using (2) in (3) gives

106 = a + (50 − 1) d

⇒ 106 = a + 49 d (5)

Subtract (4) from (5) to solve (4) and (5):

⇒ 106 − 12 = (a + 49 d) − (a + 2 d)

⇒ 94 = 47 d

∴ d = 2 (6)

Substituting value from (6) in (4) this gives:

⇒ 12 = a + 4

∴ a = 8 (7)

Substituting once more the values from (6) and (7) in (3) we obtain:

a n = 8 + 2 (n − 1) (8)

For the 29th term substitute n = 29 in (8) the result is:

a 29 = 8 + 2 (29 − 1)

⇒ a 29 = 8 + 56

∴ a 29 = 64

So, the 29th term of the A.P. is 64.

Q9. If the 3rd and the 9th terms of an A.P. are 4 and 8 respectively. Which term of this A.P. is zero.

Solution: Start with the information given in the question. the 3rd Term, a 3 = 4 (1)

Given, the 9th Term, a 9 = − 8 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Using (1) in (3) gives

4 = a + (3 − 1) d

⇒ 4 = a + 2 d (4)

Using (2) in (3) gives

− 8 = a + (9 − 1) d

⇒ − 8 = a + 8 d (5)

Subtract (4) from (5) to solve (4) and (5):

⇒ − 8 − 4 = (a + 8 d) − (a + 2 d)

⇒ − 12 = 6 d

∴ d = − 2 (6)

Substituting value from (6) in (4) this gives:

⇒ 4 = a − 4

∴ a = 8 (7)

Substituting once more the values from (6) and (7) in (3) we obtain:

a n = 8 − 2 (n − 1) (8)

For the term which is zero, substitute a n = 0 in (8)

0 = 8 − 2 (n − 1)

⇒ 8 = 2 (n − 1)

⇒ 4 = (n − 1)

∴ n = 5

So, given A.P. sequence contains 5th term as zero.

Q10. If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.

Solution: Given that the 17th term of an A.P. exceeds its 10th term by 7 i.e.,

a 17 = a 10 + 7 (1)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (2)

For 17th term substitute n = 17 in (2) i.e., a 17 = a + 16 d (3)

For 10th term substitute n = 10 in (2) i.e., a 10 = a + 9 d (4)

Hence, from (1), (3) and (4) the result is:

a + 16 d = a + 9 d + 7

⇒ 7 d = 7

∴ d = 1

So, the common difference is 1.

Q11. Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?

Solution: Let nth term of A.P. be 132 more than its 54th term i.e.,

a n = a 54 + 132 (1)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (2)

For 54th term substitute n = 54 in (2) i.e., a 54 = a + 53 d (3)

Hence, from (1), (2) and (3) we obtain:

a + (n − 1) d = a + 53 d + 132

⇒ (n − 1) d − 53 d = 132

∴ d = 132 n − 54 (4)

Now, given A.P. 3, 15, 27, 39, …

Common difference d = 15 − 3 = 12 (5)

Hence, from (4) and (5) we obtain 12 = 132 n − 54

⇒ n − 54 = 11

∴ n = 65

So, 65th term of the given A.P. will be 132 more than its 54th term.

Q12. Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?

Solution: Let 2 A.P.’s be

a, a + d, a + 2 d, a + 3 d, … (1)

b, b + d, b + 2 d, b + 3 d, … (2)

(Since common difference is same)

Given that the difference between their 100th term is 100 i.e.,

a 100 − b 100 = 100 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Hence, from (3) and (4) the result is:

a + (100 − 1) d − (b + (100 − 1) d) = 100

⇒ a − b = 100 (5)

Following the same pattern, the difference between their 1000th terms is,

a 1000 − b 1000 = [ a + (1000 − a) d ] − [ b + (1000 − a) d ]

⇒ a 1000 − b 1000 = a − b

∴ a 1000 − b 1000 = 100

So, the difference between their 1000th terms is 100.

Q13. How many three-digit numbers are divisible by 7?

Solution: First three-digit number that is divisible by 7 is 105 then the next number will be 105 + 7 = 112.

So, the series becomes 105, 112, 119, …

This arithmetic progression has first term as 105 and common difference as 7.

Now, the largest 3 digit number is 999.

Take us divide it by 7, to get the remainder.

999 = 142 × 7 + 5

So, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.

Also, this will be the last term of the A.P. series.

Hence the final sequence is as follows: 105, 112, 119, …, 994

Take 994 be the nth term of this A.P.

Then, a n = 105 + 7 (n − 1)

⇒ 994 = 105 + 7 (n − 1)

⇒ 889 = 7 (n − 1)

⇒ 127 = (n − 1)

∴ n = 128

So, 128 three-digit numbers are divisible by 7.

Q14. How many multiples of 4 lie between 10 and 250?

Solution: First number that is divisible by 4 and lie between 10 and 250 is 12. The next number will be 12 + 4 = 16.

So, the series becomes 12, 16, 20, …

This arithmetic progression has first term as 12 and common difference as 4.

Now, the largest number in range is 250.

Take us divide it by 4 to get the remainder.

250 = 62 × 4 + 2

So, 250 − 2 = 248 is the last term of the A.P. series.

Hence the final sequence is as follows: 12, 16, 20, …, 248

Take 248 be the nth term of this A.P.

Then, a n = 12 + 4 (n − 1)

⇒ 248 = 12 + 4 (n − 1)

⇒ 236 = 4 (n − 1)

⇒ 59 = (n − 1)

∴ n = 60

So, 60 multiples of 4 lie between 10 and 250.

Q15. For what value of n, are the nth terms of two APs 63, 65, 67, … and 3, 10, 17, … equal

Solution: Given 2 A.P.’s are

63, 65, 67, … (1)

Its first term is 63 and common difference is 65 − 63 = 2

3, 10, 17, … (2)

Its first term is 3 and common difference is 10 − 3 = 7

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Hence, from (1) and (3) we obtain the nth term of the first A.P. is

a n = 63 + 2 (n − 1)

⇒ a n = 61 + 2 n (4)

And from (2) and (3) we obtain the nth term of the second A.P. is

b n = 3 + 7 (n − 1)

⇒ b n = − 4 + 7 n (5)

If the nth terms of two APs 63, 65, 67, … and 3, 10, 17, … are equal the from (4) and (5),

a n = b n

⇒ 61 + 2 n = − 4 + 7 n

⇒ 65 = 5 n

∴ n = 13

So, the 13th term of both the A.P.’s are equal.

Q16. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution: Begin by writing the known quantities. The 7th term of A.P. is 12 more than its 5th term i.e.,

a 7 = a 5 + 12 (1)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (2)

For 5th term substitute n = 5 in (2) i.e., a 5 = a + 4 d (3)

For 7th term substitute n = 7 in (2) i.e., a 7 = a + 6 d (4)

Hence, from (1), (3) and (4) we obtain:

a + 6 d = a + 4 d + 12

⇒ 2 d = 12

∴ d = 6 (5)

Substituting (5) in (2) we obtain: a n = a + 6 (n − 1) (6)

Given the third term of the A.P. is 16. Hence from (6),

16 = a + 6 (3 − 1)

⇒ 16 = a + 12

∴ a = 4 (7)

Hence from (6), a n = 4 + 6 (n − 1)

So, the A.P. will be 4, 10, 16, 22, …

Q17. Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253

Solution: Given A.P. 3, 8, 13, …, 253. For the 20th term from the last write the given A.P. in reverse order and then find its 20th term.

Required A.P. is 253, …, 13, 8, 3 (1)

Its first A.P. is 253 and common difference is 8 − 13 = − 5. (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Hence from (2) and (3) we obtain: a n = 253 − 5 (n − 1) (4)

Substitute n = 20 in (4) this gives:

a 20 = 253 − 5 (20 − 1)

⇒ a 20 = 158.

So, 20th term from the last is 158.

Q18. The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.

Solution: Use the stated data to set up the progression. The sum of 4th and 8th terms of an A.P. is 24 i.e.,

a 4 + a 8 = 24 (1)

Given the sum of 6th and 10th terms is 44 i.e.,

a 6 + a 10 = 44 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

For 4th term substitute n = 4 in (3) i.e., a 4 = a + 3 d (4)

For 6th term substitute n = 6 in (3) i.e., a 6 = a + 5 d (5)

For 8th term substitute n = 8 in (3) i.e., a 8 = a + 7 d (6)

For 10th term substitute n = 10 in (3) i.e., a 10 = a + 9 d (7)

Hence, from (1), (6) and (4) we have:

(a + 3 d) + (a + 7 d) = 24

⇒ 2 a + 10 d = 24

⇒ a + 5 d = 12 (8)

From (2), (5) and (7) we have:

(a + 5 d) + (a + 9 d) = 44

⇒ 2 a + 14 d = 44

⇒ a + 7 d = 22 (9)

Subtracting (8) from (9) we have:

⇒ (a + 7 d) − (a + 5 d) = 22 − 12

⇒ 2 d = 10

∴ d = 5 (10)

Substituting this value from (10) in (9) we have:

a + 35 = 22

∴ a = − 13 (11)

Thus from (10) and (11), the first three terms of the A.P. are − 13, − 8, − 3.

Q19. Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?

Solution: Given in the first year, annual salary is ₹5000.

In the second year, annual salary is ₹5000 + 200 = 5200.

In the third year, annual salary is ₹5200 + 200 = 5400.

These values make an arithmetic progression. with first term 5000 and common difference 200.

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d

So, For the nth year, annual salary is a n = 5000 + 200 (n − 1)

⇒ a n = 4800 + 200 n …. (1)

For the year in which his annual income reaches ₹7000, substitute a n = 7000 in (1) and find the value of n i.e.,

7000 = 4800 + 200 n

⇒ 2200 + 200 n

∴ n = 11

So, in 11th year i.e., in 2005 his salary will be ₹7000.

Q20. Ramkali saved ₹5 in the first week of a year and then increased her weekly saving by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.

Solution: Given in the first week the savings is ₹5.

In the second week the savings is ₹5 + 1.75 = 6.75.

In the third week the savings is ₹6.75 + 1.75 = 8.5.

These values make an arithmetic progression. with first term 5 and common difference 1.75.

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d

So, For the nth week the savings is a n = 5 + 1.75 (n − 1)

⇒ a n = 3.25 + 1.75 n …. (1)

For the week in which her savings reaches ₹20.75, substitute a n = 20.75 in (1) and find the value of n i.e.,

20.75 = 3.25 + 1.75 n

⇒ 17.5 = 1.75 n

∴ n = 10

So, in 10th week her savings will be ₹20.75.

Exercise 5.3

Q1. Find the sum of the following APs.

(i). 2, 7, 12, … to 10 terms.

Solution: Start with the information given in the question. the first Term, a = 2 (1)

The common difference is d = 7 − 2 = 5 (2)

The term number is n = 10 (3)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (4)

Putting the known values into (4) gives S n = 10 2 [ 2 (2) + (10 − 1) (5) ]

⇒ S n = 5 [ 4 + 45 ]

∴ S n = 245

(ii). − 37, − 33, − 29, … to 12 terms

Solution: Start with the information given in the question. the first Term, a = − 37 (1)

The common difference is d = − 33 − (− 37) = 4 (2)

The term number is n = 12 (3)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (4)

Putting the known values into (4) gives S n = 12 2 [ 2 (− 37) + (12 − 1) (4) ]

⇒ S n = 6 [ − 74 + 44 ]

∴ S n = − 180

(iii). 0. 6, 1. 7, 2. 8, … to 100 terms

Solution: Start with the information given in the question. the first Term, a = 0.6 (1)

The common difference is d = 1.7 − 0.6 = 1.1 (2)

The term number is n = 100 (3)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (4)

Putting the known values into (4) gives S n = 100 2 [ 2 (0.6) + (100 − 1) (1.1) ]

⇒ S n = 50 [ 1.2 + 108.9 ]

∴ S n = 5505

(iv). 1 15, 1 12, 1 10, … to 11 terms

Solution: Start with the information given in the question. the first Term, a = 1 15 (1)

The common difference is d = 1 12 − 1 15 = 1 60 (2)

The term number is n = 11 (3)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (4)

Putting the known values into (4) gives S n = 11 2 [ 2 (1 15) + (11 − 1) (1 60) ]

⇒ S n = 11 2 [ 4 + 5 30 ]

∴ S n = 33 20

Q2. Find the sums given below

(i). 7 + 10 1 2 + 14 + … + 84

Solution: First identify the relevant values. the first Term, a = 7 (1)

The common difference is d = 10 1 2 − 7 = 7 2 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Substituting the values from (1) and (2) in (3) we obtain:

a n = 7 + 7 2 (n − 1) = 7 2 (n + 1) (4)

The final term is a n = 84 (5)

Substituting (5) in (4) we have: 84 = 7 2 (n + 1)

⇒ 24 = (n + 1)

∴ n = 23 (6)

We know that the sum of n terms of the A.P. with first term a and last term l is given by S n = n 2 [ a + l ] (7)

Putting the known values into (7) gives S n = 23 2 [ 7 + 84 ]

⇒ S n = 23 2 (91)

∴ S n = 1046 1 2

(ii). 34 + 32 + 30 + … + 10

Solution: Begin by writing the known quantities. the first Term, a = 34 (1)

The common difference is d = 32 − 34 = − 2 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Substituting the values from (1) and (2) in (3) the result is:

a n = 34 − 2 (n − 1) = 36 − 2 n (4)

The final term is a n = 10 (5)

Substituting (5) in (4) this gives: 10 = 36 − 2 n

⇒ 2 n = 26

∴ n = 13 (6)

We know that the sum of n terms of the A.P. with first term a and last term l is given by S n = n 2 [ a + l ] (7)

Putting the known values into (7) gives S n = 13 2 [ 34 + 10 ]

⇒ S n = 13 2 (44)

∴ S n = 286

(iii). − 5 + (− 8) + (− 11) + … + (− 230)

Solution: First identify the relevant values. the first Term, a = − 5 (1)

The common difference is d = − 8 − (− 5) = − 3 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Substituting the values from (1) and (2) in (3) we obtain:

a n = − 5 − 3 (n − 1) = − 2 − 3 n (4)

The final term is a n = − 230 (5)

Substituting (5) in (4) we have: − 230 = − 2 − 3 n

⇒ − 228 = − 3 n

∴ n = 76 (6)

We know that the sum of n terms of the A.P. with first term a and last term l is given by S n = n 2 [ a + l ] (7)

Putting the known values into (7) gives S n = 76 2 [ − 5 + (− 230) ]

⇒ S n = 76 2 (− 235)

∴ S n = − 8930

Q3. In an AP

(i). Given a = 5, d = 3, a n = 50, find n and S n.

Solution: Start with the information given in the question. the first Term, a = 5 (1)

The common difference is d = 3 (2)

Given, nth term of the A.P., a n = 50 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Putting the known values into (4) gives

50 = 5 + 3 (n − 1) = 2 + 3 n

After simplification,

n = 50 − 2 3

∴ n = 16 (5)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (6)

Putting the known values into (6) gives S n = 16 2 [ 2 (5) + (16 − 1) (3) ]

⇒ S n = 8 [ 10 + 45 ]

∴ S n = 440

(ii). Given a = 7, a 13 = 35, find d and S 13.

Solution: Use the stated data to set up the progression. the first Term, a = 7 (1)

Given, 13th term of the A.P., a 13 = 35 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Insert the values from (1) and (2) into (3):

35 = 7 + (13 − 1) d = 7 + 12 d

After simplification,

d = 28 12

∴ d = 7 3 (4)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (5)

Substituting the values from (1) and (4) in (5) we have: S 13 = 13 2 [ 2 (7) + (13 − 1) (7 3) ]

⇒ S 13 = 13 2 [ 14 + 28 ]

∴ S 13 = 273

(iii). Given d = 3, a 12 = 37, find a and S 12.

Solution: Begin by writing the known quantities. the common difference, d = 3 (1)

Given, 12th term of the A.P., a 12 = 37 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Insert the values from (1) and (2) into (3):

37 = a + 3 (12 − 1) = a + 33

After simplification,

∴ a = 4 (4)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (5)

Substituting the values from (1) and (4) in (5) we have: S 12 = 12 2 [ 2 (4) + (12 − 1) (3) ]

⇒ S 12 = 6 [ 8 + 33 ]

∴ S 12 = 246

(iv). Given a 3 = 1 5, S 10 = 125 find a 10 and d.

Solution: Begin by writing the known quantities. 3rd term of the A.P., a 3 = 1 5 (1)

Given, the sum of terms, S 10 = 125 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Using (1) in (3) gives

15 = a + (3 − 1) d = a + 2 d (4)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (5)

Using (1) in (5) gives 125 = 10 2 [ 2 a + (10 − 1) d ]

⇒ 125 = 5 [ 2 a + 9 d ]

∴ 25 = 2 a + 9 d (5)

Take us solve equations (4) and (5) by subtracting twice of (4) from (5) we obtain:

25 − 30 = (2 a + 9 d) − (2 a + 4 d)

⇒ − 5 = 5 d

∴ d = − 1 (6)

From (4) and (6) we obtain: a = 17 (7)

From (3), (6) and (7) for n = 10 this gives:

a 10 = 17 − (10 − 1)

∴ a 10 = 8

(v). Given S 9 = 75, d = 5 find a and a 9.

Solution: First identify the relevant values. common difference, d = 5 (1)

Given, the sum of terms, S 9 = 75 (2)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (3)

Insert the values from (1) and (2) into (3): 75 = 9 2 [ 2 a + 5 (9 − 1) ]

⇒ 25 = 3 [ a + 20 ]

⇒ 3 a = − 35

∴ a = − 35 3 (4)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (5)

Insert the values from (1) and (4) into (5):

a 9 = − 35 3 + 5 (9 − 1)

⇒ a 9 = − 35 3 + 40

∴ a 9 = 85 3

(vi) Given a = 2, d = 8, S n = 9 0, find n and a n.

Solution: Use the stated data to set up the progression. common difference, d = 8 (1)

Given, first term, a = 2 (2)

Given, the sum of terms, S n = 9 0 (3)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (3)

Substituting the values from (1), (2), (3) in (4) the result is: 90 = n 2 [ 2 (2) + 8 (n − 1) ]

⇒ 45 = n [ 2 n − 1 ]

⇒ 2 n 2 − n − 45 = 0

⇒ 2 n 2 − 10 n + 9 n − 45 = 0

⇒ 2 n (n − 5) + 9 (n − 5) = 0

⇒ (n − 5) (2 n + 9) = 0

∴ n = 5 (4)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (5)

Substituting the values from (1), (2), (4) in (5) the result is:

a 5 = 2 + 8 (5 − 1)

⇒ a 5 = 2 + 32

∴ a 5 = 34

(vii). Given a = 8, S n = 21 0, a n = 62, find n and d.

Solution: Begin by writing the known quantities. first term, a = 8 (1)

Given, the sum of terms, S n = 21 0 (2)

The stated nth term is a n = 62 (3)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (4)

Insert the values from (1) and (2) into (4): 210 = n 2 [ 2 (8) + d (n − 1) ]

⇒ 420 = n [ 16 + (n − 1) d ] (4)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (5)

Insert the values from (1) and (3) into (5):

62 = 8 + (n − 1) d (6)

Take us solve equations (4) and (6) by subtracting n times of (6) from (4) we obtain:

420 − 62 n = (16 n + n (n − 1) d) − (8 n + n (n − 1) d)

⇒ 420 − 62 n = 8 n

⇒ 420 = 70 n

∴ n = 6 (7)

Using (7) in (6) gives

62 = 8 + (6 − 1) d

⇒ 54 = 5 d

∴ d = 54 5

(viii). Given S n = − 14, d = 2, a n = 4, find n and a.

Solution: Use the stated data to set up the progression. common difference, d = 2 (1)

Given, the sum of terms, S n = − 14 (2)

The stated nth term is a n = 4 (3)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (4)

Insert the values from (1) and (2) into (4): − 14 = n 2 [ 2 a + 2 (n − 1) ]

⇒ − 14 = n [ a + n − 1 ] (5)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (6)

Insert the values from (1) and (3) into (6):

4 = a + 2 (n − 1) (7)

Take us solve equations (5) and (7) by substituting the value of a from (7) in (5) we have:

− 14 = n [ (4 − 2 (n − 1)) + n − 1 ]

⇒ − 14 = n [ 5 − n ]

⇒ n 2 − 5 n − 14 = 0

⇒ n 2 − 7 n + 2 n − 14 = 0

⇒ (n − 7) (n + 2) = 0

∴ n = 7 (Since n cannot be negative) (8)

Using (8) in (7) gives

4 = a + 2 (7 − 1)

⇒ 4 = a + 12

∴ a = − 8

(ix). Given a = 3, n = 8, S = 192, find d.

Solution: Begin by writing the known quantities. first term, a = 3 (1)

Given, the sum of terms, S n = 192 (2)

The term number is n = 8 (3)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (4)

Insert the values from (1) and (2) into (4): 192 = 8 2 [ 2 (3) + d (8 − 1) ]

⇒ 192 = 4 [ 6 + 7 d ]

⇒ 48 = 6 + 7 d

⇒ 42 = 7 d

∴ d = 6

(x). Given l = 28, S = 144 and there are total 9 terms. Find a.

Solution: First identify the relevant values. last term, l = 28 (1)

Given, the sum of terms, S n = 144 (2)

The term number is n = 9 (3)

We know that the sum of n terms of the A.P. with first term a and last term l is given by S n = n 2 [ a + l ] (4)

Insert the values from (1) and (2) into (4): 144 = 9 2 [ a + 28 ]

⇒ 32 = a + 28

∴ a = 4

Q4. How many terms of the A.P. 9, 17, 25… must be taken to give a sum of 636?

Solution: First identify the relevant values. common difference, d = 17 − 9 = 8 (1)

Given, first term, a = 9 (2)

Given, the sum of terms, S n = 636 (3)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (3)

Substituting the values from (1), (2), (3) in (4) this gives: 636 = n 2 [ 2 (9) + 8 (n − 1) ]

⇒ 636 = n (5 + 4 n)

⇒ 4 n 2 + 5 n − 636 = 0

⇒ 4 n 2 + 53 n − 48 n − 636 = 0

⇒ n (4 n + 53) − 12 (4 n + 53) = 0

⇒ (n − 12) (4 n + 53) = 0

⇒ n = 12

o r

− 53 4

Since n can only be a natural number ∴ n = 12

Q5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution: Start with the information given in the question. first term, a = 5 (1)

Given, the sum of terms, S n = 400 (2)

The stated nth term is a n = 45 (3)

We know that the sum of n terms of the A.P. with first term a and last term l is given by S n = n 2 [ a + l ] (4)

Substituting the values from (1), (2), (3) in (4) we obtain: 400 = n 2 [ 5 + 45 ]

⇒ 400 = 25 n

∴ n = 16

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (5)

Insert the values from (1) and (3) into (5):

45 = 5 + (16 − 1) d

⇒ 40 = 15 d

∴ d = 8 3

Q6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution: First identify the relevant values. first term, a = 17 (1)

The common difference is d = 9 (2)

The stated nth term is a n = 350 (3)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (4)

Substituting the values from (1), (2), (3) in (4) this gives:

350 = 17 + 9 (n − 1)

⇒ 333 = 9 (n − 1)

⇒ 37 = (n − 1)

∴ n = 38 (5)

We know that the sum of n terms of the A.P. with first term a and last term l is given by S n = n 2 [ a + l ] (6)

Substituting the values from (1), (5), (3) in (6) we have: S 38 = 38 2 [ 17 + 350 ]

⇒ S 38 = 19 (367)

∴ S 38 = 6973

Q7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Solution: Begin by writing the known quantities. the common difference, d = 7 (1)

Given, the 22nd term, a 22 = 149 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Insert the values from (1) and (2) into (3):

149 = a + 7 (22 − 1)

⇒ 149 = a + 147

∴ a = 2 (4)

We know that the sum of n terms of the A.P. with first term a and last term l is given by S n = n 2 [ a + l ] (5)

Substituting the values from (1), (2), (4) in (5) we have: S 22 = 22 2 [ 2 + 149 ]

⇒ S 22 = 11 (151)

∴ S 22 = 1661

Q8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution: Begin by writing the known quantities. the 2nd term, a 2 = 14 (1)

Given, the 3rd term, a 3 = 18 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d (3)

Using (1) in (3) gives

14 = a + d (4)

Using (2) in (3) gives

18 = a + 2 d (5)

Subtract (4) from (5) to solve (4) and (5):

18 − 14 = (a + 2 d) − (a + d)

∴ d = 4 (6)

Substituting the value from (6) in (4) this gives a = 10. (7)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (8)

Substituting the values from (7), (6) in (8) we obtain for n = 51,

S 51 = 51 2 [ 2 (10) + 4 (51 − 1) ]

⇒ S 51 = 51 2 [ 20 + 200 ]

∴ S 51 = 5610

Q9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Solution: Start with the information given in the question. the sum of first 7 terms, S 7 = 49 (1)

Given, the sum of first 17 terms, S 17 = 289 (2)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (3)

Using (1) in (3) gives

49 = 7 2 [ 2 a + (7 − 1) d ]

⇒ 7 = a + 3 d (4)

Using (2) in (3) gives

289 = 17 2 [ 2 a + (17 − 1) d ]

⇒ 17 = a + 8 d (5)

Subtract (4) from (5) to solve (4) and (5):

17 − 7 = (a + 8 d) − (a + 3 d)

⇒ 10 = 5 d

∴ d = 2 (6)

Substituting the value from (6) in (4) we obtain a = 1. (7)

Insert the values from (7) and (6) into (3):

S n = n 2 [ 2 + 2 (n − 1) ]

∴ S n = n 2

Q10. Show that a 1, a 2 …, a n, … form an AP where a n is defined as below. Also find the sum of the first 15 terms in each case.

(i). a n = 3 + 4 n

Solution: Consider two consecutive terms of the given sequence. Say a n, a n + 1. Difference between these terms will be

a n + 1 − a n = [ 3 + 4 (n + 1) ] − [ 3 + 4 n ]

⇒ a n + 1 − a n = 4 (n + 1) − 4 n

⇒ a n + 1 − a n = 4

Which is a constant ∀ n ∈ N.

For n = 1, a 1 = 3 + 4 = 7

So, it is an A.P. with first term 7 and common difference 4.

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ]

So, S 15 = 15 2 [ 2 (7) + 4 (15 − 1) ]

⇒ S 15 = 15 2 [ 14 (5) ]

∴ S 15 = 525

(ii). a n = 9 − 5 n

Solution: Consider two consecutive terms of the given sequence. Say a n, a n + 1. Difference between these terms will be

a n + 1 − a n = [ 9 − 5 (n + 1) ] − [ 9 − 5 n ]

⇒ a n + 1 − a n = − 5 (n + 1) + 5 n

⇒ a n + 1 − a n = − 5

Which is a constant ∀ n ∈ N.

For n = 1, a 1 = 9 − 5 = 4

So, it is an A.P. with first term 4 and common difference − 5.

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ]

So, S 15 = 15 2 [ 2 (4) − 5 (15 − 1) ]

⇒ S 15 = 15 [ − 31 ]

∴ S 15 = − 465

Q11. If the sum of the first n terms of an AP is 4 n − n 2, what is the first term (that is S 1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the 10th and the nth terms.

Solution: Begin by writing the known quantities. the sum of the first n terms of an A.P. is 4 n − n 2.

First term = S 1 = 4 − 1 = 3. (1)

Sum of first two terms = S 2 = 8 − (2) 2 = 4 (2)

From (1) and (2), 2nd term = S 2 − S 1 = 4 − 3 = 1.

Sum of first three terms = S 3 = 12 − (3) 2 = 3 (3)

From (3) and (2), 3rd term = S 3 − S 2 = 3 − 4 = − 1.

Following the same pattern,

Sum of first n terms = S n = 4 n − n 2 (4)

Sum of first n − 1 terms = S n − 1 = 4 (n − 1) − (n − 1) 2 = − n 2 + 6 n − 5 (5)

From (4) and (5), nth term = S n − S n − 1 = (4 n − n 2) − (− n 2 + 6 n − 5) = 5 − 2 n (6)

From (6), 10th term is 5 − 2 (10) = − 15.

Q12. Find the sum of first 40 positive integers divisible by 6.

Solution: First positive integer that is divisible by 6 is 6 itself.

Second positive integer that is divisible by 6 is 6 + 6 = 12.

Third positive integer that is divisible by 6 is 12 + 6 = 18.

Hence, it is an A.P. with first term and common difference both as 6.

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ]

So, for n = 40,

S 40 = 40 2 [ 2 (6) + 6 (40 − 1) ]

⇒ S 40 = 120 [ 41 ]

∴ S 40 = 4920

Q13. Find the sum of first 15 multiples of 8.

Solution: First positive integer that is divisible by 8 is 8 itself.

Second positive integer that is divisible by 8 is 8 + 8 = 16.

Third positive integer that is divisible by 8 is 16 + 8 = 24.

Hence, it is an A.P. with first term and common difference both as 8.

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ]

So, for n = 15,

S 15 = 15 2 [ 2 (8) + 8 (15 − 1) ]

⇒ S 15 = 60 [ 16 ]

∴ S 15 = 960

Q14. Find the sum of the odd numbers between 0 and 50.

Solution: The odd numbers between 0 and 50 are 1, 3, 5, …, 49.

It is an A.P. with first term 1 and common difference 2. ….(1)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d … (2)

Substitute a n = 49 and values from (1) into (2)

49 = 1 + 2 (n − 1)

⇒ 24 = (n − 1)

∴ n = 25 (3)

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ] (4)

Insert values from (1) and (3) into (4):

S 25 = 25 2 [ 2 + 2 (25 − 1) ]

⇒ S 25 = 25 [ 25 ]

∴ S 25 = 625

Q15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

Solution: Penalty of delay for first day is ₹200.

Penalty of delay for second day is ₹250.

Penalty of delay for third day is ₹300.

Hence it is an A.P. with first term 200 and common difference 50.

Money the contractor has to pay as penalty, if he has delayed the work by 30 days is the sum of first 30 terms of the A.P.

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ]. So,

S 30 = 30 2 [ 2 (200) + 50 (30 − 1) ]

⇒ S 30 = 15 [ 400 + 50 (29) ]

∴ S 30 = 27750

So, the contractor has to pay ₹27750 as penalty.

Q16. 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: Let the first prize be of ₹a then the second prize will be of ₹a − 20, the third prize will be of ₹a − 40.

So, it is an A.P. with first term a and common difference − 20.

Given, S 7 = 700

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ]. So,

S 7 = 7 2 [ 2 a − 20 (7 − 1) ]

⇒ 700 = 7 [ a − 60 ]

⇒ 100 = a − 60

∴ a = 160

So, the value of each of the prizes was ₹

160,

₹

140,

₹

120, ₹

100,

₹

80,

₹

60,

a n d

₹

40.

Q17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

Solution: Each section of class I will plant 1 tree each. So, total trees planted by class I are 3.

Each section of class II will plant 2 trees each. So, total trees planted by class II are 3 × 2 = 6.

Each section of class III will plant 3 trees each. So, total trees planted by class III are 3 × 3 = 9.

So, it is an A.P. sequence with first term and common difference both as 3.

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ]. So,

S 12 = 12 2 [ 2 (3) − 3 (12 − 1) ]

⇒ S 12 = 6 [ 39 ]

∴ S 12 = 234

So, 234 trees will be planted by the students.

Q18. A spiral is made up of successive semicircles, with centers alternately at A and B, starting with center at A of radii 0. 5, 1. 0 cm, 1. 5 cm, 2. 0 cm, … . What is the total length of such a spiral made up of thirteen consecutive semicircles?

a spiral made up of thirteen consecutive semicircles

Solution: Length of first semi-circle I 1 = π (0.5) cm.

Length of second semi-circle I 2 = π (1) cm.

Length of third semi-circle I 3 = π (1.5) cm.

So, it is an A.P. sequence with first term and common difference both as π (0.5).

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ]. So,

S 13 = 13 2 [ 2 (0.5 π) + (0.5 π) (13 − 1) ]

⇒ S 13 = 7 × 13 × (0.5 π)

⇒ S 13 = 7 × 13 × 1 2 × 22 7

∴ S 13 = 143

So, the length of such spiral of thirteen consecutive semi-circles

will be 143 cm.

Q19. The 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

how many logs are in the top row

Solution: Total logs in first row are 20.

Total logs in second row are 19.

Total logs in third row are 18.

So, it is an A.P. sequence with first term 20 and common difference − 1.

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ]. So,

200 = n 2 [ 2 (20) − (n − 1) ]

⇒ 400 = n [ 41 − n ]

⇒ n 2 − 41 n + 400 = 0

⇒ n 2 − 16 n − 25 n + 400 = 0

⇒ n (n − 16) − 25 (n − 16) = 0

⇒ (n − 16) (n − 25) = 0

For n = 25, after 20th term, all terms are negative, which is illogical as terms are representing the number of logs and number of logs being negative is illogical.

∴ n = 16

Total logs in 16th row = 20 − (16 − 1) = 5

So, 200 logs will be placed in 16 rows and the total logs in 16th row will be 5.

Q20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

(Hint: to pick up the first potato and the second potato, the total

distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3))

the total distance the competitor has to run

Solution: Total distance run by competitor to collect and drop first potato = 2 × 5 = 10 m.

Total distance run by competitor to collect and drop second potato = 2 × (5 + 3) = 16 m.

Total distance run by competitor to collect and drop third potato = 2 × (5 + 3 + 3) = 22 m.

So, it is an A.P. sequence with first term 10 and common difference 6.

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ]. So, to collect and drop 10 potatoes total distance covered is

S 10 = 10 2 [ 2 (10) + 6 (10 − 1) ]

⇒ S 10 = 5 [ 74 ]

∴ S 13 = 370

So, the competitor will run a total distance of 370 m.

Exercise 5.4

Q1. Which term of the A.P. 121, 117, 113, … is its first negative term?

(Hint: Find n for a n < 0)

Solution: Given A.P. 121, 117, 113, …

Its first term is 121 and common difference is 117 − 121 = − 4.

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d.

Hence the nth term of the given A.P. is a n = 121 − 4 (n − 1) (1)

To find negative term, find n such that a n < 0

Hence from (1),

121 − 4 (n − 1) < 0

⇒ 121 < 4 (n − 1)

⇒ 121 4 + 1 < n

⇒ n > 125 4

∴ n > 31.25

So, the 32nd term of the given A.P. will be its first negative term.

Q2. The sum of the third and the seventh terms of an A.P is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.

Solution: Begin by writing the known quantities. The sum of third and seventh term of A.P., a 3 + a 7 = 6 (1)

Given the sum of third and seventh term of A.P., a 3 ⋅ a 7 = 8 (2)

Use the nth-term formula for an arithmetic progression: a n = a + (n − 1) d. So,

For n = 3, a 3 = a + 2 d

For n = 7, a 7 = a + 6 d

From (1), a 3 + a 7 = (a + 2 d) + (a + 6 d)

⇒ 2 a + 8 d = 6

∴ a + 4 d = 3 (3)

From (2), a 3 ⋅ a 7 = (a + 2 d) ⋅ (a + 6 d)

∴ a 2 + 8 a d + 12 d 2 = 8 (4)

Take us now solve equations (3) and (4) by substituting the value of a from (3) into (4).

(3 − 4 d) 2 + 8 d (3 − 4 d) + 12 d 2 = 8

⇒ 9 − 24 d + 16 d 2 + 24 d − 32 d 2 + 12 d 2 = 8

⇒ − 4 d 2 + 1 = 0

⇒ d 2 = 1 4

∴ d = 1 2, − 1 2 (5)

CASE 1: For d = 1 2

Substitute d = 1 2 in (6) we obtain: a = 1 (6)

So, it is an A.P. sequence with first term 1 and common difference 1 2.

Use the sum formula for the first n terms of an arithmetic progression: S n = n 2 [ 2 a + (n − 1) d ]. So,

S 16 = 16 2 [ 2 + 1 2 (16 − 1) ]

⇒ S 16 = 4 [ 19 ]

∴ S 16 = 76

CASE 2: For d = − 1 2

Substitute d = − 1 2 in (6) we have: a = 5 (7)

So, it is an A.P. sequence with first term 5 and common difference − 1 2 and hence,

S n = n 2 [ 2 a + (n − 1) d ]

⇒ S 16 = 16 2 [ 2 (5) − 1 2 (16 − 1) ]

⇒ S 16 = 4 [ 5 ]

∴ S 16 = 20

Q3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2 1 2 m apart, what is the length of the wood required for the rungs?

(Hint: number of rungs = 250 25)

the length of the wood required for the rungs

Solution: Distance between first and last rungs is 2 1 2 m = 5 2 m = 250 c m.

Distance between two consecutive rungs is 25 c m.

So, total number of rungs are 250 25 + 1 = 11.

Also, Notice that the length of each rung is decreasing in a uniform order. So, we can conclude that the length of rungs is in A.P. with first term 45, common difference − 25 and number of terms 11.

We know that the sum of n terms of the A.P. with first term a and last term l is given by S n = n 2 [ a + l ]. So,

S 11 = 11 2 [ 45 + 25 ]

⇒ S 11 = 11 [ 35 ]

∴ S 11 = 385

So, the length of the wood required for the rungs is 385 cm.

Q4. The houses of a row are number consecutively from 1 to 49. Show that there is a value of x such that the sum of numbers of the houses preceding the house numbered x is equal to the sum of the number of houses following it.

Find this value of x.

(Hint: S_{x − 1} = S₄₉ − S_x)

Solution: Given houses are numbered 1, 2, 3, 4, …

Clearly, they are numbered in A.P. series with both first term and common difference as 1.

Now, there is house numbered x such that the sum of numbers of the houses preceding the house numbered x is equal to the sum of the number of houses following it i.e., S_{x − 1} = S₄₉ − S_x

the sum of the number of houses

We know that the sum of n terms of the A.P. with first term a and last term l is given by S_n = n/2 [ a + l ]. So,

S_{x − 1} = S₄₉ − S_x

⇒ { (x − 1)/2 [ 1 + (x − 1) ] } = { 49/2 [ 1 + 49 ] } − { x/2 [ 1 + x ] }

⇒ x (x − 1)/2 = 49 [ 25 ] − x (x + 1) 2

⇒ x (x − 1) = 2450 − x (x + 1)

⇒ 2x² = 2450

∴ x = 35 (Since house number cannot be negative)

So, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.

Q5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m calculate the total volume of concrete required to build the terrace. The total volume of concrete required to build the terrace

Solution: Given that a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. An easy illustration of the problem is depicted below.

An easy illustration of the problem is depicted

Here blue step is the lowermost step. Let it be known as step 1. The volume of step 1 is 1/2 × 1/4 × 50 m³.

The red step is the second lowermost step. Let it be known as step 2. The volume of step 2 is 1/2 × 1/2 × 50 m3.

The green step is the third lower step. Let it be known as step 3. The volume of step 3 is 1/2 × 1 × 50 m³.

the height is increasing with each increasing

We can see that the height is increasing with each increasing step by a factor of 1 4, length and width being constant. Hence the volume of each step is increasing by 1/2 × 1/4 × 50 m³.

Thus, the volume of steps is in A.P. with first term and common difference both as 1/2 × 1/4 × 50 = 25/4 m³.

Use the sum formula for the first n terms of an arithmetic progression: S_n = n/2 [ 2 a + (n − 1) d ]. So,

S₁₅ = 15/2 [ 2 (25 4) + (25 4) (15 − 1) ]

⇒ S₁₅ = 15/2 (25 4) [ 16 ]

⇒ S₁₅ = 15 ⋅ 25 ⋅ 2

∴ S₁₅ = 750

So, volume of concrete required to build the terrace is 750 m³.