**‘NCERT Solutions for Class 10 Maths Chapter**

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## NCERT Class 10 Maths Textbook Chapter 5 With Answer Book PDF Free Download

### Chapter 5: Arithmetic Progressions

#### 5.1 Introduction

You must have observed that in nature, many things follow a certain pattern, such as the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the spirals on pineapple and on a pine cone, etc.

We now look for some patterns which occur in our day-to-day life. Some such examples are :

(i) Reena applied for a job and got selected. She has been offered a job with a starting monthly salary of 8000, with an annual increment of 500 in her salary. Her salary (in `) for the 1st, 2nd, 3rd, . . . years will be, respectively 8000, 8500, 9000, . . . .

(ii) The lengths of the rungs of a ladder decrease uniformly by 2 cm from bottom to top (see Fig. 5.1). The bottom rung is 45 cm in length. The lengths (in cm) of the 1st, 2nd, 3rd, . . ., 8th rung from the bottom to the top are, respectively 45, 43, 41, 39, 37, 35, 33, 31

Author | NCERT |

Language | English |

No. of Pages | 24 |

PDF Size | 1729 KB |

Category | Mathematics |

Source/ Credits | ncert.nic.in |

### NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions

1. In which of the following situations, does the list of numbers involved make an arithmetic progression and why?

**(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.**

**Solution:**

We can write the given condition as;

Taxi fare for 1 km = 15

Taxi fare for first 2 kms = 15+8 = 23

Taxi fare for first 3 kms = 23+8 = 31

Taxi fare for first 4 kms = 31+8 = 39

And so on……

Thus, 15, 23, 31, 39 … forms an A.P. because every next term is 8 more than the preceding term.

**(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.**

**Solution:**

Let the volume of air in a cylinder, initially, be *V* liters.

In each stroke, the vacuum pump removes 1/4th of air remaining in the cylinder at a time. Or we can say, after every stroke, 1-1/4 = 3/4th part of air will remain.

Therefore, volumes will be *V*, 3*V*/4, (3*V*/4)^{2}, (3*V*/4)^{3}…and so on

Clearly, we can see here, the adjacent terms of this series do not have a common difference between them. Therefore, this series is not an A.P.

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

**Solution:**

We can write the given condition as;

Cost of digging a well for first metre = Rs.150

Cost of digging a well for first 2 metres = Rs.150+50 = Rs.200

Cost of digging a well for first 3 metres = Rs.200+50 = Rs.250

Cost of digging a well for first 4 metres =Rs.250+50 = Rs.300

And so on..

Clearly, 150, 200, 250, 300 … forms an A.P. with a common difference of 50 between each term.

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

**Solution:**

We know that if Rs. P is deposited at *r*% compound interest per annum for n years, the amount of money will be:

P(1+r/100)^{n}

Therefore, after each year, the amount of money will be;

10000(1+8/100), 10000(1+8/100)^{2}, 10000(1+8/100)^{3}……

Clearly, the terms of this series do not have a common difference between them. Therefore, this is not an A.P.

**2. Write the first four terms of the A.P. when the first term a and the common difference are given as follows**:

**(i) a = 10, d = 10**

(ii) a = -2, d = 0

(iii) a = 4, d = – 3

(iv) a = -1 d = 1/2

(v) a = – 1.25, d = – 0.25

**Solutions:**

(i) *a* = 10, *d* = 10

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{1} = *a* = 10

*a*_{2} = *a*_{1}+*d* = 10+10 = 20

*a*_{3} = *a*_{2}+*d* = 20+10 = 30

*a*_{4} = *a*_{3}+*d* = 30+10 = 40

*a*_{5} = *a*_{4}+*d* = 40+10 = 50

And so on…

Therefore, the A.P. series will be 10, 20, 30, 40, 50 …

And First four terms of this A.P. will be 10, 20, 30, and 40.

(ii) *a* = – 2, *d* = 0

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{1} = *a* = -2

*a*_{2} = *a*_{1}+*d* = – 2+0 = – 2

*a*_{3} = *a*_{2}+d = – 2+0 = – 2

*a*_{4} = *a*_{3}+*d* = – 2+0 = – 2

Therefore, the A.P. series will be – 2, – 2, – 2, – 2 …

And, First four terms of this A.P. will be – 2, – 2, – 2 and – 2.

(iii) *a* = 4, *d* = – 3

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{1} = *a* = 4

*a*_{2} = *a*_{1}+*d* = 4-3 = 1

*a*_{3} = *a*_{2}+*d* = 1-3 = – 2

*a*_{4} = *a*_{3}+*d* = -2-3 = – 5

Therefore, the A.P. series will be 4, 1, – 2 – 5 …

And, first four terms of this A.P. will be 4, 1, – 2 and – 5.

(iv) *a* = – 1, *d* = 1/2

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{2} = *a*_{1}+*d* = -1+1/2 = -1/2

*a*_{3} = *a*_{2}+*d* = -1/2+1/2 = 0

*a*_{4} = *a*_{3}+*d* = 0+1/2 = 1/2

Thus, the A.P. series will be-1, -1/2, 0, 1/2

And First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v) *a* = – 1.25, *d* = – 0.25

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{1} = *a* = – 1.25

*a*_{2} = *a*_{1} + *d* = – 1.25-0.25 = – 1.50

*a*_{3} = *a*_{2} + *d* = – 1.50-0.25 = – 1.75

*a*_{4} = *a*_{3} + *d* = – 1.75-0.25 = – 2.00

Therefore, the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 ……..

And first four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.

**3. For the following A.P.s, write the first term and the common difference.(i) 3, 1, – 1, – 3 …(ii) -5, – 1, 3, 7 …(iii) 1/3, 5/3, 9/3, 13/3 ….(iv) 0.6, 1.7, 2.8, 3.9 …**

**Solutions**

(i) Given series,

3, 1, – 1, – 3 …

First term, *a* = 3

Common difference, *d* = Second term – First term

⇒ 1 – 3 = -2

⇒ d = -2

**(ii) Given series, – 5, – 1, 3, 7 …**

First term, *a* = -5

Common difference, *d* = Second term – First term

⇒ ( – 1)-( – 5) = – 1+5 = 4

**(iii) Given series, 1/3, 5/3, 9/3, 13/3 ….**

First term, *a* = 1/3

Common difference, *d* = Second term – First term

⇒ 5/3 – 1/3 = 4/3

**(iv) Given series, 0.6, 1.7, 2.8, 3.9 …**

First term, *a* = 0.6

Common difference, *d* = Second term – First term

⇒ 1.7 – 0.6

⇒ 1.1

NCERT Class 10 Maths Textbook Chapter 5 With Answer Book PDF Free Download