NCERT Solutions For Class 10 Math Chapter – 5 Exercise – 5.1

NCERT  Solutions For Class 10 Math Chapter – 5 Exercise – 5.1

The exercise solutions provided here are created with a vision to assist the students in their first term exam preparation.The  Exercise solution cover all the questions given in Exercise 5.1 of the NCERT textbook. These solutions are researched and prepared by our subject experts. Studying this study material will aid you in solving different questions on distance from a point.

Exercise solutions are provided in PDF format for your easy access and download. By studying this solution you can clear your doubts on distance calculation. By studying this study material you can get well acquainted with all important formulas. Here, you can also get knowledge of alternative methods to solve distance from a point problems.

 

 

Q1.

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

  • The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional k.

Solutions:-

 

It can be observed that

Taxi fare for 1st Km  = 15

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

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

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

 

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

 

  • 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 initial volume of air in a cylinder be V liters In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time. In other words , after every strokes , only 1-1/4 = 3/4th part of air will remain. Therefore , volumes will be (3v/4) (3V/4)2 (3V/4)3

Clearly , it can be observed that the adjacent terms of this series do not have the same difference between them Therefore this is not an A.P.

 

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

Solution:-

 

Cost of digging for first meter = 150

Cost of digging for first 2 meters = 150 + 50 = 200

Cost of digging for first 3 meters = 200 + 50 = 250

Cost of digging for first 4 meter = 250 + 50 = 300

Clearly , 150,200,250,300….. forms an AP because every term is 50 more than the preceding term.

 

  • The amount of money in the account every year , when Rs 10000 is deposited at compounded interest at 8% per annum.

Solution:-

 

We know that if Rs P is deposited at  r% compound interest per annum for n year our money will be

P(1 + r/100)n

Therefore  after every year , our  money will be

10000(1 + 8/100) ,10000 (1 + 8/100)2  , 10000(1 + 8/100)3

 

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

 

Q2. Write first four terms of the A.P. when the first term a and the common difference are given as follows.

  • A = 10  , d = 10
  • A = -2   , d = 10
  • A  = 4  , d = -3
  • A= -1  d = 1/2
  • A = 1.25    d = -0.25

Solution:-

 

  • a = 10  d=10

Let the series be a1 , a2 , a3 , a4 , a5 …

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

 

  • A = -2  d = 0

Let the series be a1 , a2 , a3  , a4 …..

Therefore the series will be -2 , -2 , -2 , -2 ……

 

  • A = 4   d = -3

Let the series be a1 , a2 , a3 , a4 .….

Therefore the series will be 4 , 1, -2 , -5

 

  • A = -1  , d = 1/2

Let the series be a1 , a2 , a3 , a4 ….

Therefore the series will be -1  , -1/2 , 0  , 1/2

 

  • A = -1.25  d = 0.25

 

Let the series be a1 , a2  , a3 , a4 ….

Clearly the series will be 1.25  , -1.50 , -1.75  , -2.00 ….

 

Q3. For the following A.Ps write the first term and the common difference  

  • 3 , 1 , -1 , -3….
  • -5 , -1 , 3 , 7 …..
  • 1/3,5/3,9/3,13/3…..
  • 6,1.7,2.8,3.9

Solution:-

 

  • 3 , 1 , -1 , -3…

Here first term a = 3

Common difference d = Second term – First term

= 1 – 3 = 2

 

  • -5 , -1 , 3 , 7

A = -5

D = 4

  • 1/3 , 5/3 , 9/3 , 13/3

A = 1/3

D = 4/3

 

  • 6 , 1.7 , 2.8 , 3.9

A = 0.6

D = 1.1

 

Q4. Which of the following are APs ? If they form an A.P. find the common difference d and write three more terms.

  • 2 , 4 , 8 , 16….
  • 2 , 5/2 , 3 , 7/2 ….
  • -1.2 , -3.2 , -5.2 , -7.2 …..
  • -10 , -6 , -2 , 2….
  • 3 , 3 + √2 , 3 + 2√2 , 3 + 3√2
  • 2 , 0.22 , 0.222 , 0.2222…
  • 0 , -4 , -8 , -12….
  • -1/2 , -1/2 , -1/2 , -1/2 …
  • 1,3,9,27
  • A,2a,3a,4a
  • A,a2,a3,a4
  • 2 , 8 , 18 , 32….
  • 3,6,9,12….
  • 12, 32 , 52  , 72
  • 12, 52 , 72 , 73 …..

Solution:-

 

  • 2,4,8,16 …..

Here

A2 – a1 = 4 – 2 = 2

A3 – a2 = 8 – 4 = 4

A4 – a3 = 16 – 8 = 8

Therefore the given number are forming an A.P.

 

  • 2 , 5/2 , 3 , 7/2

Here ,

a2 – a1 = 5/2 – 2 = 1/2

A3 – a2 = 3 – 5/2 = 1/2

A4 – a3 = 7/2 – 3 = 1/2

A5 = 4

A6 = 9/2

a7 = 5

 

 

  • -1.2 , -3.2 , -5.2 , -7.2

A2 – a1 = (-3.2) – (-1.2) = -2

A3 – a2 = (-5.2) – (-3.2)  = -2

A4 – a3 = (-7.2) – (5.2) = -2

 

A5 = -9.2

A6 = -11.2

A7 = -13.2

 

  • -10 , -6 , -2 , 2 ….

A2 – a1 = (-6) – (-10) = 4

A3 – a2 = (-2) – (-6) = 4

A4 – a3= (2) – (-2) = 4

A5 = 6

A6 = 10

A7 = 14

 

  • 3 , 3+√2 , 3 +2√2 , 3 + 3√2

A2 – a1 = √2.

A3 -a2= √2

A4 – a3 = √2.

A5 = 3 +4√2

A6 = 3 + 5√2

A7 = 3 + 6√2.

 

  • 2 , 0.22 , 0.222 , 0.2222….

a2 – a1 = 0.22 –  0.2 = 0.02

A3 – a2 = 0.222 – 0.22 = 0.002

A4 – a3 =    0.2222 – 0.222 =  0.0002.

 

  • 0 , -4 , -8 , -12 ……

A2 – a1 = (-4) – 0 = -4

A3 – a2 = (-8) – (-4) = -4

A4 – a3 = (-12) – (-8) = -4

 

A5 = -12 – 4 = -16

A6 = -16 – 4 = -20

A7 = -20  – 4 = -24

 

  • -1/2  , -1/2  , -1/2 , -1/2

A2 – a1 =  (-1/2) – (-1/2) = 0

A3 – a2 = (-1/2) – (-1/2) = 0

A4 – a3 = (-1/2) – (-1/2) = 0

A5 =  -1/2

A6 = -1/2

A7 = -1/2

 

  • 1 , 3 , 9 , 27

A2 – a1 = 3 – 1 = 2

A3 – a2 = 9 – 3 = 6

2 ≠ 6.

 

  • A , 2a , 3a , 4….

A2 – a1 = 2a – a = a

A3 – a2 = 3a – 2a = a

A4 – a3 = 4a – 3a = a

A5 = 5a

A6 = 6a

A7 = 7a

 

  • A , a2, a3 , a4 …..

A2 – a1 = a2 – a = (a-1)

A3 – a2 = a3 – a2 = a2 (a-1)

A4 – a3 = a4 – a3 = a3 (a-1)

 

  • √2 , √8 , √18 , √32……

a2 – a1 = √8 – √2 = 2√2 – √2 = √2

A3 – a1 = √18  – √8 = √2

A4 – a3 = √32 – √18 = √2

A5 = √50

A6 = √72

A7 = √98

 

 

  • √3 , √6 , √9 , √12…..

A2 – a1 =  √6 – √3 = √3(√2 – 1)

A3 – a2 = √9 – √3 = √3(√3 – √2)

A4 – a3 = √12 – √9 = √3 (2-√3)

 

  • 12, 32 , 52 , 72…..

A2 – a1 = 9 – 1 = 8

A3  – a2 = 25 – 9 = 16

 

  • 12, 32 ,  52 , 72 , ….

A2 – a1 = 25 – 1 = 24

A3 – a2 = 49 – 25 = 24

A4 – a3 = 73 – 49 = 24

A5 = 97

A6 = 121

A7 = 145