NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Arithmetic Progressions |

Exercise |
Ex 5.1, Ex 5.2, Ex 5.3 |

Number of Questions Solved |
4 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

**Question 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 ₹ 15 for the first km and ₹ 8 for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes \(\frac { 1 }{ 4 }\) of the air remaining in the cylinder at a time.

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

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

**Solution:**

(i) Given:

a_{1} = ₹ 15

a_{2} = ₹ 15 + ₹ 8 = ₹ 23

a_{3} = ₹ 23 + ₹ 8 = ₹ 31

List of fares is ₹ 15, ₹ 23, ₹ 31

and a_{2} – a_{1} = ₹ 23 – ₹ 15 = ₹ 8

a_{3} – a_{2} = ₹ 31 – ₹ 23 = ₹ 8

Here, a_{2} – a_{1} = a_{3} – a_{2}

Thus, the list of fares forms an AP.

(iii) Given:

a_{1} = ₹ 150, a_{2} = ₹ 200, a_{3} = ₹ 250

a_{2} – a_{1} = ₹ 200 – ₹ 150 = ₹ 50

and a_{3} – a_{2} = ₹ 250 – ₹ 200 = ₹ 50

Here, a_{3} – a_{2} = a_{2} – a_{1}

Thus, the list forms an AP.

(iv) Given: a_{1} = ₹ 10000

a_{2} = ₹ 10000 + ₹ 10000 x \(\frac { 8 }{ 100 }\) = ₹ 10000 + ₹ 800 = ₹ 10800

a_{3} = ₹ 10800 + ₹ 10800 x \(\frac { 8 }{ 100 }\) = ₹ 10800 + ₹ 864 = ₹ 11664

a_{2} – a_{1} = ₹ 10800 – ₹ 10000 = ₹ 800

a_{3} – a_{2} = ₹ 11664 – ₹ 10800 = ₹ 864

a_{3} – a_{2} ≠ a_{2} – a_{1}

Thus, it is not an AP.

**Question 2.**

Write first four terms of the AP, when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10

(n) a = -2, d = 0

(iii) a = 4, d = -3

(iv) a = -1, d = \(\frac { 1 }{ 2 }\)

(v) a = -1.25, d = -0.25

**Solution:**

(i) Given: a = 10, d = 10

a_{1} = 10,

a_{2} = 10 + 10 = 20

a_{3} = 20 + 10 = 30

a_{4} = 30 + 10 = 40

Thus, the first four terms of the AP are 10, 20, 30, 40.

(ii) Given: a = – 2, d = 0

The first four terms of the AP are -2, -2, -2, -2.

(iii) a_{1} = 4, d = -3

a_{2} = a_{1} + d = 4 – 3 = 1

a_{3} = a_{2} + d = 1 – 3 = -2

a_{4} = a_{3} + d = -2 – 3 = -5

Thus, the first four terms of the AP are 4, 1, -2, -5.

(iv)

(v) a_{1} = -1.25, d = -0.25

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

Thus, the first four terms of the AP are -1.25, -1.50, -1.75, -2.

**Question 3.**

For the following APs, write the first term and the common difference:

(i) 3, 1, -1, -3, ……

(ii) -5, -1, 3, 7, ……

(iii) \(\frac { 1 }{ 3 }\) , \(\frac { 5 }{ 3 }\) , \(\frac { 9 }{ 3 }\), \(\frac { 13 }{ 3 }\) , ……..

(iv) 0.6, 1.7, 2.8, 3.9, …….

**Solution:**

(i) a_{1} = 3, a_{2} = 1

d = a_{2} – a_{1} = 1 – 3 = -2

where, a_{1} = first term and d = common difference

a_{1} = 3, d = -2

(ii) a_{1} = -5, a_{2} = -1

d = a_{2} – a_{1} = -1 – (-5) = -1 + 5 = 4

So, first term a_{1} = -5 and common difference d = 4

(iii) a_{1} = \(\frac { 1 }{ 3 }\), a_{2} = \(\frac { 5 }{ 3 }\)

d = \(\frac { 5 }{ 3 }\) – \(\frac { 1 }{ 3 }\) = \(\frac { 4 }{ 3 }\)

So, first term a_{1} = \(\frac { 1 }{ 3 }\) and common difference d = \(\frac { 4 }{ 3 }\)

a_{1}= 0.6, a_{2} = 1.7

d = a_{2} – a_{1} = 1.7 – 0.6 = 1.1

So, first term a_{1} = 0.6 and common difference d = 1.1

**Question 4.**

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

(i) 2, 4, 8, 16, …….

(ii) 2, \(\frac { 5 }{ 2 }\) , 3, \(\frac { 7 }{ 2 }\) , …….

(iii) -1.2, -3.2, -5.2, -7.2, ……

(iv) -10, -6, -2,2, …..

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …..

(vi) 0.2, 0.22, 0.222, 0.2222, ……

(vii) 0, -4, -8, -12, …..

(viii) \(\frac { -1 }{ 2 }\) , \(\frac { -1 }{ 2 }\) , \(\frac { -1 }{ 2 }\) , \(\frac { -1 }{ 2 }\) , …….

(ix) 1, 3, 9, 27, …….

(x) a, 2a, 3a, 4a, …….

(xi) a, a2, a3, a4, …….

(xii) √2, √8, √18, √32, …..

(xiii) √3, √6, √9, √12, …..

(xiv) 12, 32, 52, 72, ……

(xv) 12, 52, 72, 73, ……

**Solution:**

(i) 2, 4, 8, 16, ……

a_{2} – a_{1} = 4 – 2 = 2

a_{3} – a_{2} = 8 – 4 = 4

a_{2} – a_{1} ≠ a_{3} – a_{2}

Thus, the given sequence is not an AP.

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