Ex 5.1 class 10 CBSE complete solution

Looking for the Maths NCERT Ex 5.1 class 10 full in detailed solution? You came to the right place. We have solved each problems of Ex 5.1 class 10 in this pagucatione and explained the reason behind it. Hope you will find it helpful in solving your maths exercise.

Ex 5.1 class 10
Ex 5.1 class 10

Ex 5.1 class 10 CBSE complete solution

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.

(ii) The amount of air present in a cylinder when a vacuum pump removes 14th of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every meter of digging, when it costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter.

(iv) The amount of money in the account every year, when Rs 10,000 is deposited at compound Interest at 8% per annum.

Ans. (i) Taxi fare for 1st km = Rs 15, Taxi fare after 2 km = 15 + 8 = Rs 23

Taxi fare after 3 km = 23 + 8 = Rs 31

Taxi fare after 4 km = 31 + 8 = Rs 39

Therefore, the sequence is 15, 23, 31, 39…

It is an arithmetic progression because difference between any two consecutive terms is equal which is 8. (23 – 15 = 8, 31 – 23 = 8, 39 – 31 = 8, …)

(ii)Let amount of air initially present in a cylinder = V

Amount of air left after pumping out air by vacuum pump =

Amount of air left when vacuum pump again pumps out air

=

So, the sequence we get is like

Checking for difference between consecutive terms …

 

Difference between consecutive terms is not equal.

Therefore, it is not an arithmetic progression.

(iii) Cost of digging 1 meter of well = Rs 150

Cost of digging 2 meters of well = 150 + 50 = Rs 200

Cost of digging 3 meters of well = 200 + 50 = Rs 250
Therefore, we get a sequence of the form 150, 200, 250…
It is an arithmetic progression because difference between any two consecutive terms is equal. (200 – 150 = 250 – 200 = 50…)

Here, difference between any two consecutive terms which is also called common difference is equal to 50.

(iv)Amount in bank after Ist year = … (1)

Amount in bank after two years = … (2)

Amount in bank after three years = … (3)

Amount in bank after four years = … (4)

It is not an arithmetic progression because (2) − (1) ≠ (3) − (2)

(Difference between consecutive terms is not equal)

Therefore, it is not an Arithmetic Progression.

Ex 5.1 class 10

2. Write first four terms of the AP, when the first term a and common difference d are given as follows:
(i)a = 10, d = 10

(ii) a = -2, d = 0

(iii) a = 4, d = -3

(iv) a = -1, d =

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

Ans. (i) First term = a = 10, d = 10

Second term = a + d = 10 + 10 = 20

Third term = second term + d = 20 + 10 = 30

Fourth term = third term + d = 30 + 10 = 40

Therefore, first four terms are: 10, 20, 30, 40

(ii) First term = a = –2 , d = 0

Second term = a + d = –2 + 0 = –2

Third term = second term + d = –2 + 0 = –2

Fourth term = third term + d = –2 + 0 = –2

Therefore, first four terms are: –2, –2, –2, –2

(iii) First term = a = 4, d =–3

Second term = a + d = 4 – 3 = 1

Third term = second term + d = 1 – 3 = –2

Fourth term = third term + d = –2 – 3 = –5

Therefore, first four terms are: 4, 1, –2, –5

(iv) First term = a = –1, d =

Second term = a + d = –1 + = −

Third term = second term + d = −+ = 0

Fourth term = third term + d = 0 + =

Therefore, first four terms are: –1, −, 0,

(v) First term = a = –1.25, d = –0.25

Second term = a + d = –1.25 – 0.25 = –1.50

Third term = second term + d = –1.50 – 0.25 = –1.75

Fourth term = third term + d

= –1.75 – 0.25 = –2.00

Therefore, first four terms are: –1.25, –1.50, –1.75, –2.00

Read about FIR from here

Ex 5.1 class 10

3. For the following APs, write the first term and the common difference.
(i) 3, 1, –1, –3 …

(ii) –5, –1, 3, 7…

(iii)

(iv) 0.6, 1.7, 2.8, 3.9 …

Ans. (i) 3, 1, –1, –3…

First term = a = 3,

Common difference (d) = Second term – first term = Third term – second term and so on

Therefore, Common difference (d) = 1 – 3 = –2

NCERT Solutions for Class 10 Maths Exercise 5.1

(ii) –5, –1, 3, 7…

First term = a = –5

Common difference (d) = Second term – First term

= Third term – Second term and so on

Therefore, Common difference (d) = –1 – (–5) = –1 + 5 = 4

Ex 5.1 class 10

(iii)

First term = a =

Common difference (d) = Second term – First term

= Third term – Second term and so on

Therefore, Common difference (d) =

(iv) 0.6, 1.7, 2.8, 3.9…

First term = a = 0.6

Common difference (d) = Second term – First term

= Third term – Second term and so on

Therefore, Common difference (d) = 1.7 − 0.6 = 1.1

Ex 5.1 class 10

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, , 3, …

(iii) −1.2, −3.2, −5.2, −7.2…

(iv) −10, −6, −2, 2…

(v)

(vi) 0.2, 0.22, 0.222, 0.2222…

(vii) 0, −4, −8, −12…

(viii)

(ix) 1, 3, 9, 27…

(x) a, 2a, 3a, 4a…

(xi)

(xii)

(xiii)

(xiv)

(xv)

Ans. (i) 2, 4, 8, 16…

It is not an AP because difference between consecutive terms is not equal.

As4 – 2 ≠ 8 − 4

(ii)2, , 3, …

It is an AP because difference between consecutive terms is equal.

 

Common difference (d) =

Fifth term = Sixth term = 4 + ½ =

Seventh term =

Therefore, next three terms are 4, and 5.

Ex 5.1 class 10

(iii)−1.2, −3.2, −5.2, −7.2…

It is an AP because difference between consecutive terms is equal.

−3.2 − (−1.2)

= −5.2 − (−3.2)

= −7.2 − (−5.2) = −2

Common difference (d) = −2

Fifth term = −7.2 – 2 = −9.2Sixth term = −9.2 – 2 = −11.2

Seventh term = −11.2 – 2 = −13.2

Therefore, next three terms are −9.2, −11.2 and −13.2

(iv) −10, −6, −2, 2…

It isan AP because difference between consecutive terms is equal.

−6 − (−10) = −2 − (−6)

= 2 − (−2) = 4

Common difference (d) = 4

Fifth term = 2 + 4 = 6 Sixth term = 6 + 4 = 10

Seventh term = 10 + 4 = 14

Therefore, next three terms are 6, 10 and 14

Ex 5.1 class 10

(v)

It is an AP because difference between consecutive terms is equal.

Common difference (d) =

Fifth term =

Sixth term =

Seventh term =

Therefore, next three terms are

(vi) 0.2, 0.22, 0.222, 0.2222…

It is not an AP because difference between consecutive terms is not equal.

0.22 − 0.2 ≠ 0.222 − 0.22

(vii) 0, −4, −8, −12…

It is an AP because difference between consecutive terms is equal.

−4 – 0 = −8 − (−4)

= −12 − (−8) = −4

Common difference (d) = −4

Fifth term = −12 – 4 =−16 Sixth term = −16 – 4 = −20

Seventh term = −20 – 4 = −24

Therefore, next three terms are −16, −20 and −24

Ex 5.1 class 10

(viii)

It is an AP because difference between consecutive terms is equal.

 

Common difference (d) = 0

Fifth term = Sixth term =

Seventh term =

Therefore, next three terms are

(ix) 1, 3, 9, 27…

It is not an AP because difference between consecutive terms is not equal.

3 – 1 ≠ 9 − 3

(x) a, 2a, 3a, 4a…

It is an AP because difference between consecutive terms is equal.

2a – a = 3a − 2a = 4a − 3a = a

Common difference (d) = a

Fifth term = 4a + a = 5a Sixth term = 5a + a = 6a

Seventh term = 6a + a = 7a

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

(xi) a, a2, a3, a4…

It is not an AP because difference between consecutive terms is not equal.

a2 – a ≠ a3 − a2

Ex 5.1 class 10

(xii)

 

It is an AP because difference between consecutive terms is equal.

 

 

Common difference (d) =

Fifth term = Sixth term =

Seventh term =

Therefore, next three terms are

(xiii)

It is not an AP because difference between consecutive terms is not equal.

 

(xiv)

It is not an AP because difference between consecutive terms is not equal.

Ex 5.1 class 10

(xv)

1, 25, 49, 73…

It is an AP because difference between consecutive terms is equal.

 

= = 24

Common difference (d) = 24

Fifth term = 73 + 24 = 97 Sixth term = 97 + 24 = 121

Seventh term = 121 + 24 = 145

Therefore, next three terms are 97, 121 and 145

 

Ex 5.1 class 10 : Things to remember in Arithmetic Progression (AP)

Consider
(i) 1, 2, 3, 4, ……
(ii) 3, 3, 3, 3, …..
(i) and (ii) are the sequence of numbers, each number in these sequences is called a term.

An arithmetic progression (AP) is a sequence of numbers in which each term is obtained by adding a fixed number ‘d’ to the preceeding term, except the first term.
The fixed number is called the common difference. It can be positive, negative or zero.
(Ex 5.1 class 10)Any Arithmetic progression can be represented as :
a, a + d, a + 2d, a + 3d,…..
where ‘a’ is the first term and ‘d’ is the common difference. Arithmetic progressions which does not have a last term are called Infinite Arithmetic Progression. e.g.:
6, 9, 12, 15,…….

Formula for common Difference (d)
A sequence of numbers a1, a2, a3…. is an AP if the difference a2 – a1, a3 – a2, a4 – a3…. gives the same value, i.e. if ak+1 – ak is the same for different values of k. The difference (ak+1 – ak) is called common difference (d). Here ak+1 and ak are the (k + 1)th and kth terms respectively.
∴ d = a2 – a1 = a3 – a2 = a4 – a3

nth Term (or General Term) of an Arithmetic Progressions
In an AP, with first term ‘a’ and common difference d, the nth term(or the general term) is given by,
an = a + (n – 1)d
Note that an AP can be finite or infinite according to as the number of terms are finite or infinite.
If there are m terms in an AP then am is the last term and is sometimes denoted by ‘l’.

Sum of the FIRST ‘n’ Terms of an A.P.
(i) The sum of the first n terms of an A.P. is given by
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 Q1
where a is the first term and d is the common difference
(ii) If l is the last term of the finite A.P. say the nth term, then the sum of all terms of the A.P. is given by,
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 Q2
Note that sum of first n positive integers is given by
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 Q3

Arithmetic Mean Between Two Numbers(Ex 5.1 class 10)
If a, b, c are in AP. Then b is called the arithmetic mean of a and c and is given by
.

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