NCERT Solutions for Arithmetic Progressions Chapter 5 Ex 5.1 Class 10 Maths has all the solutions to the questions provided in the NCERT Book of the latest edition.

Students are advised to practice all the questions to get good marks in the board examination.

Textbook | NCERT |

Class | 10 |

Subject | Mathematics |

Chapter | 5 |

Exercise | 5.1 |

Chapter Name | Arithmetic Progressions |

## Class 10 Maths Chapter 5 Exercise 5.1 Arithmetic Progressions NCERT Solution

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

Answer

Yes, because $$15,23,31,…$$ forms an A.P. as each term is obtained by adding 8 in its preceding term.

(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.

Answer

Let the amount of air present be V

According to question, after removing $$\frac{1}{4}$$ of air remaining in the cylinder

the air present is $$\frac{3}{4}$$

So,

the air present are $$V, \frac{3}{4}V, \frac{3}{4}\times \frac{3}{4}V, \frac{3}{4}\times \frac{3}{4}\times \frac{3}{4}V$$

$$=V, \frac{3}{4}V, \left(\frac{3}{4}\right)^{2}V, \left(\frac{3}{4}\right)^{3}V, ….$$

$$\therefore$$ No, it is not A.P. because the common difference is not same

(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.

Answer

Yes, because $$150, 200,250,300,…$$ forms an A.P. as each term is obtained by adding 50 to its preceding term.

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

Answer

As we know, $$A=P(1+r)^{t}$$ where

A = Amount, P = Principal, r = interest rate, t = time

Here,

P = ₹10000, r = 8%

$$\therefore$$ Amount for consecutive years are

$$10000\left(1+\frac{8}{100}\right)^{1}, 10000\left(1+\frac{8}{100}\right)^{2}, 10000\left(1+\frac{8}{100}\right)^{3},…. $$

So, it is not an A.P. because difference is not same.

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

Answer

$$\because$$ First four terms of A.P. can be written as $$a, a+d, a+2d, a+3d$$

$$\therefore AP=10,20,30,40$$

(ii) a = –2, d = 0

Answer

First four terms of A.P. can be written as $$a, a+d, a+2d, a+3d$$

$$\therefore AP=-2,-2,-2,-2$$

(iii) a = 4, d = – 3

Answer

First four terms of A.P. can be written as $$a, a+d, a+2d, a+3d$$

$$\therefore AP=4,1,-2,-5$$

(iv) a = – 1, d = $$\frac{1}{2}$$

Answer

First four terms of A.P. can be written as $$a, a+d, a+2d, a+3d$$

$$\therefore AP=-1,\frac{-1}{2}, 0, \frac{1}{2}$$

(v) a = – 1.25, d = – 0.25

Answer

$$\because$$ First four terms of A.P. can be written as $$a, a+d, a+2d, a+3d$$

$$\therefore AP=-1.25, -1.50, -1.75, -2.0$$

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

(i) 3, 1, – 1, – 3, . . .

Answer

a = 3, d = $$1-3=-2$$

(ii) – 5, – 1, 3, 7, . . .

Answer

a = -5, d = $$-1-(-5)=4$$

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

Answer

a = $$\frac{1}{3}$$, d = $$\frac{5}{3}-\frac{1}{3}=\frac{4}{3}$$

(iv) 0.6, 1.7, 2.8, 3.9, . . .

Answer

a = 0.6, d = $$1.7-0.6=1.1$$

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, . . ..

Answer

For A.P., the given sequence should satisfy this relation 2b = a + c

$$\because 2b= 2\times 4 =8$$

a + c = 8 + 2 = 10

$$\because$$ This sequence doesn’t satisfy the relation

$$\therefore$$ It is not an A.P.

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

Answer

$$\because 2b= 2\times \frac{5}{2} =5$$

a + c = 2 + 3 = 5

$$\because$$ This sequence satisfy the relation 2b = a + c

$$\therefore$$ It is an A.P.

So,

d = $$\frac{5}{2}-{2}={1}{2}$$

and

next three terms are $$4,\frac{9}{2},5$$

(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . .

Answer

$$\because 2b= 2\times (-3.2) = -6.4$$

a + c = -1.2 + (-5.2) = -6.4

$$\because$$ This sequence satisfy the relation 2b = a + c

$$\therefore$$ It is an A.P.

So,

d = $$-3.2 – (-1.2) = -2.0$$

and

next three terms are $$-9.2, -11.2, -13.2$$

(iv) – 10, – 6, – 2, 2, . . .

Answer

$$\because 2b= -12$$, a + c = -12

$$\therefore$$ It is an A.P.

So, d = $$-6-(-10)=4$$

next three terms are 6, 10, 14

(v) $$3, 3 +\sqrt{2} , 3+ 2\sqrt{2}, 3+ 3 \sqrt{2}, . . .$$

Answer

$$\because 2b= 6+2\sqrt{3}$$, a + c = $$6+2\sqrt{2}$$

$$\therefore$$ It is an A.P.

So, d = $$3+\sqrt{2}-3=\sqrt{2}$$

next three terms are $$3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}$$

(vi) 0.2, 0.22, 0.222, 0.2222, . . .

Answer

$$\because 2b= 0.44$$, a + c = 0.422

$$\therefore$$ It is not an A.P.

(vii) 0, – 4, – 8, –12, . . .

Answer

$$\because 2b= -8$$, a + c = -8

$$\therefore$$ It is an A.P.

So,

d = – 4 – 0 = – 4$$

next three terms are $$-16, -20, -24$$

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

Answer

$$\because 2b= -1$$, a + c = -1

$$\therefore$$ It is an A.P.

So,

d = $$-\frac{1}{2}-\left(-\frac{1}{2}\right) = 0$$

next three terms are $$-\frac{1}{2}, -\frac{1}{2},-\frac{1}{2}$$

(ix) 1, 3, 9, 27, . . .

Answer

$$\because 2b=6$$, a + c = 10

$$\therefore$$ It is not an A.P.

(x) a, 2a, 3a, 4a, …

Answer

$$\because 2b=4a$$, a + c = 4a

$$\therefore$$ It is an A.P.

So, d = 2a – a = a

next three terms are 5a, 6a, 7a

(xi) $$a, a^{2},a^{3},a^{4}, …$$

Answer

$$\because 2b=2a^{2}$$, a + c = $$a^{3} + a$$

$$\therefore$$ It is not an A.P.

(xii) $$\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32}, …$$

Answer

$$\because 2b=2\sqrt{8}$$

$$=2\times \sqrt{4\times 2}$$

$$=4\sqrt{2}$$

a + c = $$\sqrt{2}+\sqrt{18}$$

$$=\sqrt{2}+3\sqrt{2}$$

$$=4\sqrt{2}$$

$$\therefore$$ It is an A.P.

and

AP = $$\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, …$$

So, d = $$2\sqrt{2}-\sqrt{2} = \sqrt{2}$$

next three terms are $$5\sqrt{2}, 6\sqrt{2}, 7\sqrt{2}$$

$$= \sqrt{50}, \sqrt{72}, \sqrt{98}$$

(xiii) $$\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},…$$

Answer

$$\because 2b=2\sqrt{6}$$, a + c = $$\sqrt{3} + \sqrt{9} = 3+\sqrt{3}$$

$$\therefore$$ It is not an A.P.

(xiv) $$1^{2},3^{2},5^{2},7^{2},…$$

Answer

$$\because 2b=2\times 9=18$$, a + c = 5^{2} + 1 = 26

$$\therefore$$ It is not an A.P.

(xv) $$1^{2},5^{2},7^{2},73,…$$

Answer

$$\because 2b=25\times 2=50$$, a + c = 1 + 49 = 50

$$\therefore$$ It is an A.P.

So, $$d=5^{2}-1=24$$

next three terms are 97, 121, 145

Hope NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1, helps you in solving problems. If you have any doubts, drop a comment below and we will get back to you.