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

### (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 = Ï€R2H

Amount of air removed by vacuum pump = Ï€R2H/4

Remaining air = 3 * Ï€R2H/4

Again amount of air removed by vacuum pump = 3 * Ï€R2H/16

Remaining air = 3 * Ï€R2H/4 – 3 * Ï€R2H/16 = 9 * Ï€R2H/16

and so on

The series can be written as: Ï€R2H, 3 * Ï€R2H/4, 9*Ï€R2H/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.

### (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,….

### (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

### (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 â‰  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+âˆš2, 3+2âˆš2, 3+3âˆš2,….

Solution:

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

d1 = a2 + a1 = 3 + âˆš2 – 3 = âˆš2

d2 = a3 + a2 = 3 + 2âˆš2 – (3 + âˆš2) = âˆš2

So d1 = d2

Hence, the above series is in arithmetic progression and the common difference is âˆš2

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

Fifth term(a5) = a4 + d = (3 + 3âˆš2) + âˆš2 = 3 + 4âˆš2

Sixth term(a6) = a5 + d = (3 + 4âˆš2) + âˆš2 = 3 + 5âˆš2

Seventh term(a7) = a6 + d = (3 + 5âˆš2) + âˆš2 = 3 + 6âˆš2

So, the next three terms are 3+4âˆš2, 3+5âˆš2, 3+6âˆš2

### (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 â‰  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 â‰  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 â‰  d2

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

### (xii). âˆš2, âˆš8, âˆš18, âˆš32,….

Solution:

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

d1 = a2 + a1 = 2âˆš2 – âˆš2 = âˆš2

d2 = a3 + a2 = 3âˆš2 – 2âˆš2 = âˆš2

So d1 = d2

Hence, the above series is in arithmetic progression and the common difference is âˆš2

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

Fifth term(a5) = a4 + d = âˆš32 + âˆš2 = 5âˆš2

Sixth term(a6) = a5 + d = 5âˆš2 + âˆš2 = 6âˆš2

Seventh term(a7) = a6 + d = 6âˆš2 + âˆš2 = 7âˆš2

So, the next three terms are 5âˆš2, 6âˆš2, 7âˆš2

### (xiii). âˆš3, âˆš6, âˆš9, âˆš12,….

Solution:

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

d1 = a2 + a1 = âˆš6 – âˆš3 = âˆš3(âˆš2 – 1)

d2 = a3 + a2 = âˆš9 – âˆš6 = âˆš3(âˆš3 – âˆš2)

So d1 â‰  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 â‰  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

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