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Class 10 NCERT Solutions- Chapter 5 Arithmetic Progressions – Exercise 5.1

  • Last Updated : 03 Mar, 2021

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 Rs. 15 for the first km and Rs. 8 for each additional km.

Solution:  

Initially fare = 0

After completing 1 km, fare = 15

After completing 2 km, fare = 15 + 8 = 23

After completing 3 km, fare = 23 + 8 = 31



And so on

So you can write fare in series as

15, 23, 31, 39,……… and so on

Here you can see that the first term is 15 and the difference between any two-term is 8. So, this series is in arithmetic progression.

Note: A series is said to be in arithmetic progression if any term can be found out by adding a fixed number to the previous term. The first term is denoted as and the fixed number is known as a difference which is denoted by d.

In the above series a = 15 and d = 8

(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: 

Initial volume of air in cylinder = πR2H



Amount of air removed by vacuum pump = πR2H/4

Remaining air = 3 * πR2H/4

Again amount of air removed by vacuum pump = 3 * πR2H/16

Remaining air = 3 * πR2H/4 – 3 * πR2H/16 = 9 * πR2H/16 

and so on

The series can be written as: πR2H, 3 * πR2H/4, 9*πR2H/16,……… and so on

You can see here is no common difference between any two-term. So, this series is not arithmetic progression.

(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.

Solution: 

Cost of digging well for first meter is 150 rupees

Cost of digging well for two meters is 150 + 50 = 200 rupees



Cost of digging well for three meters is 200 + 50 = 250 rupees

And so on

The series for cost of digging a well look like 150, 200, 250, 300,……… and so on

Here you can see that after the first term, any term can be found out by adding 50 to the previous term. Hence, the above series is in arithmetic progression with the first term 150 and the common difference is 50.

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

Solution:  

Formula for compound interest is given by P(1 + r/100)n where P is the principal amount, r is the rate of interest and n is the time duration.

So, the series can be written as: 10000(1 + 8/100), 10000(1 + 8/100)2,10000(1 + 8/100)3,……… and so on

As time duration increases, compound interest will increase exponentially. Hence, this can’t form a series in arithmetic series because there is no fixed common difference between any two terms. 

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

(i). a = 10, d = 10

Solution: 

Given:



a1 = a = 10

d = 10

Now we find the remaining terms:

Second term(a2) = a1 + d = 10 + 10 = 20

Third term(a3) = a2 + d = 20 + 10 = 30

Fourth term(a4) = a3 + d = 30 + 10 = 40

and so on.

So, four terms of this A.P. will be as follows

10, 20, 30, 40,….

(ii) a = -2, d = 0

Solution: 

Given:

a1 = a = -2

d = 0

Now we find the remaining terms:

Second term(a2) = a1 + d = -2 + 0 = -2

Third term(a3) = a2 + d = -2 + 0 = -2

Fourth term(a4) = a3 + d = -2 + 0 = -2

and so on.

So, four terms of this A.P. will be as follows

-2, -2, -2, -2,….

(iii). a = 4, d = -3

Solution: 

Given:

a1 = a = 4

d = -3

Now we find the remaining terms:

Second term(a2) = a1 + d = 4 + (-3) = 1

Third term(a3) = a2 + d = 1 + (-3) = -2

Fourth term(a4) = a3 + d = -2 + (-3) = -5

and so on.

So, four terms of this A.P. will be as follows

4, 1, -2, -5,….

(iv). a = -1, d = 1/2 

Solution: 

Given:

a1 = a = -1

d = 1/2

Now we find the remaining terms:

Second term(a2) = a1 + d = -1 + 1/2 = -1/2

Third term(a3) = a2 + d = (-1/2) + (1/2) = 0

Fourth term(a4) = a3 + d = 0 + (1/2) = 1/2

and so on.



So, four terms of this A.P. will be as follows

-1, -1/2, 0, 1/2,….

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

Solution: 

Given:

a1 = a = -1.25

d = -0.25

Now we find the remaining terms:

Second term(a2) = a1 + d = -1.25 – 0.25 = – 1.50

Third term(a3) = a2 + d = -1.50 – 0.25 = -1.75

Fourth term(a4) = a3 + d = -1.75 – 0.25 = -2.00

and so on.

So, four terms of this A.P. will be as follows

-1.25, -1.50, -1.75, -2.00,….

Question 3: For the following A.P.s, write the first term and the common difference.

(i). 3, 1, -1, -3,….

Solution: 

From the above A.P., first term (a) = 3

Common difference (d) = second term – first term

= 1 – 3 = -2

(ii). -5, -1, 3, 7,….

Solution: 

From the above A.P., first term (a) = -5

Common difference (d) = second term – first term

= -1 – (-5)

= -1 + 5 = 4

(iii). 1/3, 5/3, 9/3, 13/3,….

Solution: 

From the above A.P., first term (a) = 1

Common difference (d) = second term – first term

= 5/3 – 1/3

= 4/3

(iv). 0.6, 1.7, 2.8, 3.9,….

Solution:

From the above A.P., first term (a) = 0.6

Common difference (d) = second term – first term 

= 1.7 – 0.6

= 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,….

Solution: 

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 4 – 2 = 2

d2 = a3 + a2 = 8 – 4 = 4

So d1 ≠ d2 

Hence, this series doesn’t form an AP because there is no fixed common difference.

(ii). 2, 5/2, 3, 7/2,….

Solution: 

First we check the given series is AP or not by finding common difference:



d1 = a2 + a1 = 5/2 – 2 = 1/2

d2 = a3 + a2 = 3 – 5/2 = 1/2

So d1 = d2 

Hence, the above series is in arithmetic progression and the common difference is 1/2.

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = 7/2 + 1/2 = 4

Sixth term(a6) = a5 + d = (4) + (1/2) = 9/2

Seventh term(a7) = a6 + d = 9/2 + (1/2) = 5

So, the next three terms are: 4, 9/2, 5

(iii). -1.2, -3.2, -5.2, -7.2,….

Solution: 

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = -3.2 – (-1.2) = -2.0

d2 = a3 + a2 = -5.2 – (-3.2) = -2.0

So d1 = d2 

Hence, the above series is in arithmetic progression and the common difference is -2.0

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = (-7.2) + (-2.0) = -9.2

Sixth term(a6) = a5 + d = (-9.2) + (-2.0) = -11.2

Seventh term(a7) = a6 + d = (-11.2) + (-2.0) = -13.2

So, the next three terms are: -9.2, -11.2, -13.2

(iv). -10, -6, -2, 2,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = -6 – (-10) = 4

d2 = a3 + a2 = -2 – (-6) = 4

So d1 = d2 

Hence, the above series is in arithmetic progression and the common difference is 4

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = 2 + 4 = 6

Sixth term(a6) = a5 + d = 6 + 4 = 10

Seventh term(a7) = a6 + d = 10 + 4 = 14

So, the next three terms are 6, 10, 14

(v). 3, 3+√2, 3+2√2, 3+3√2,….

Solution: 

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 3 + √2 – 3 = √2

d2 = a3 + a2 = 3 + 2√2 – (3 + √2) = √2

So d1 = d2 

Hence, the above series is in arithmetic progression and the common difference is √2

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = (3 + 3√2) + √2 = 3 + 4√2

Sixth term(a6) = a5 + d = (3 + 4√2) + √2 = 3 + 5√2



Seventh term(a7) = a6 + d = (3 + 5√2) + √2 = 3 + 6√2

So, the next three terms are 3+4√2, 3+5√2, 3+6√2

(vi). 0.2, 0.22, 0.222, 0.2222,….

Solution: 

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 0.22 – 0.2 = 0.02

d2 = a3 + a2 = 0.222 – 0.22 = 0.002

So d1 ≠ d2 

Hence, this series doesn’t form an AP because there is no fixed common difference.

(vii). 0, -4, -8, -12,….

Solution: 

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = -4 – 0 = -4

d2 = a3 + a2 = -8 – (-4) = -4

So d1 = d2 

Hence, the above series is in arithmetic progression and the common difference is -4

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = (-12) + (-4) = -16

Sixth term(a6) = a5 + d = (-16) + (-4) = -20

Seventh term(a7) = a6 + d = (-20) + (-4) = -24

So, the next three terms are -16, -20, -24

(viii). -1/2, -1/2, -1/2, -1/2,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = -1/2 – (-1/2) = 0 

d2 = a3 + a2 = -1/2 – (-1/2) = 0 

So d1 = d2 

Hence, the above series is in arithmetic progression and the common difference is 0

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = (-1/2) + 0 = -1/2

Sixth term(a6) = a5 + d = (-1/2) + 0 = -1/2

Seventh term(a7) = a6 + d = (-1/2) + 0 = -1/2

So, the next three terms are -1/2, -1/2, -1/2

(ix). 1, 3, 9, 27,….

Solution: 

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 3 – 1 = 2

d2 = a3 + a2 = 9 – 3 = 6

So d1 ≠ d2 

Hence, this series doesn’t form an AP because there is no fixed common difference.

(x). a, 2a, 3a, 4a,….

Solution: 

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 2a – a = a

d2 = a3 + a2 = 3a – 2a = a



So d1 = d2 

Hence, the above series is in arithmetic progression and the common difference is a

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = 4a + a = 5a

Sixth term(a6) = a5 + d = 5a + a = 6a

Seventh term(a7) = a6 + d = 6a + a = 7a

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

(xi). a, a2, a3, a4,….

Solution: 

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = a2 – a 

d2 = a3 + a2 = a3 – a2  

So d1 ≠ d2 

Hence, this series doesn’t form an AP because there is no fixed common difference.

(xii). √2, √8, √18, √32,….

Solution: 

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 2√2 – √2 = √2

d2 = a3 + a2 = 3√2 – 2√2 = √2

So d1 = d2 

Hence, the above series is in arithmetic progression and the common difference is √2

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = √32 + √2 = 5√2

Sixth term(a6) = a5 + d = 5√2 + √2 = 6√2

Seventh term(a7) = a6 + d = 6√2 + √2 = 7√2

So, the next three terms are 5√2, 6√2, 7√2

(xiii). √3, √6, √9, √12,….

Solution: 

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = √6 – √3 = √3(√2 – 1)

d2 = a3 + a2 = √9 – √6 = √3(√3 – √2)

So d1 ≠ d2 

Hence, this series doesn’t form an AP because there is no fixed common difference.

(xiv). 12, 32, 52, 72,….

Solution: 

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 9 – 1 = 8

d2 = a3 + a2 = 25 – 9 = 16

So d1 ≠ d2 

Hence, this series doesn’t form an AP because there is no fixed common difference.

(xv). 12, 52, 72, 73,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 25 – 1 = 24

d2 = a3 + a2 = 49 – 25 = 24

So d1 = d2 

Hence, the above series is in arithmetic progression and the common difference is 24

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = 73 + 24 = 97

Sixth term(a6) = a5 + d = 97 + 24 = 121

Seventh term(a7) = a6 + d = 121 + 24 = 145

So, the next three terms are 97, 121, 145

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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