NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

NCERT Solutions Class 10 Maths
Chapter – 5 (Arithmetic Progressions)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 5 Arithmetic Progressions Exercise 5.1 has been provided here to help the students in solving the questions from this exercise.

Chapter : 5 Arithmetic Progressions

Exercise – 5.1

1. In which of the following situations, does the list of numbers involved make as 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 – 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
Thus, 15, 23, 31, 39 … forms an A.P. because every next term is 8 more than the preceding term.

(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 – Let the volume of air in a cylinder, initially, be V litres.
In each stroke, the vacuum pump removes 1/4th of air remaining in the cylinder at a time. Or we can say, after every stroke, 1 – 1/4 = 3/4th part of air will remain.
Therefore, volumes will be V, 3V/4 , (3V/4)² , (3V/4)³… and so on
Clearly, we can see here, the adjacent terms of this series do not have the common difference between them. Therefore, this series is not an A.P.

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

Solution – Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 … forms an A.P. with a common difference of 50 between each term.

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

Solution – We know that if Rs. P is deposited at r% compound interest per annum for n years, the amount of money will be: P(1 + r/100)ⁿTherefore, after each year, the amount of money will be;
10000(1 + 8/100), 10000(1 + 8/100)², 10000(1 + 8/100)³……
Clearly, the terms of this series do not have the common difference between them. Therefore, this is not an A.P.

2. Write first four terms of the A.P. when the first term a and the common difference are given as follows:
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = – 3
(iv) a = -1 d = 1/2
(v) a = – 1.25, d = – 0.25

Solutions –

The general form of an arithmetic progression is a, (a + d), (a + 2d), (a + 3d), … where a is the first term and d is a common difference.

(i) a = 10, d = 10
First term, a = 10
Second term, a + d = 10 + 10 = 20
Third term, a + 2d = 10 + 20 = 30
Fourth term, a + 3d = 10 + 30 = 40
The first four terms of the AP are 10, 20, 30, and 40.

(ii) a = – 2, d = 0
First term, a = – 2
Second term, a + d = – 2 + 0 = – 2
Third term, a + 2d = – 2 + 0 = – 2
Fourth term, a + 3d = – 2 + 0 = – 2
The first four terms of the AP are – 2, – 2, – 2, and -2.

(iii) a = 4, d = – 3
First term, a = 4
Second term, a + d = 4 + (- 3) = 1
Third term, a + 2d = 4 + 2(- 3) = 4 – 6 = – 2
Fourth term, a + 3d = 4 + 3(- 3) = 4 – 9 = – 5
The first four terms of the AP are 4, 1, – 2, – 5.

(iv) a = – 1, d = 1/2
First term, a = – 1
Second term, a + d = – 1 + 1/2 = – 1/2
Third term, a + 2d = – 1 + 2(1/2) = – 1 + 1 = 0
Fourth term, a + 3d = – 1 + 3(1/2) = – 1 + 3/2 = 1/2
The first four terms of the AP are – 1, – 1/2, 0, 1/2

(v) a = – 1.25, d = – 0.25
First term, a = – 1.25
Second term, a + d = -1.25 + (-0.25) = – 1.25 – 0.25 = – 1.5
Third term, a + 2d = – 1.25 + 2 (- 0.25) = – 1.25 – 0.50 = – 1.75
Fourth term, a + 3d = – 1.25 + 3 (-0.25) = – 1.25 – 0.75 = – 2
The first four terms of the AP are – 1.25, – 1.5, – 1.75, and – 2.00.

3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, – 1, – 3 …
(ii) -5, – 1, 3, 7 …
(iii) 1/3, 5/3, 9/3, 13/3 ….
(iv) 0.6, 1.7, 2.8, 3.9 …

Solutions –

The general form of an arithmetic progression is a, (a + d), (a + 2d), (a + 3d),… where a is the first term and d is a common difference.

(i) 3, 1, – 1, – 3 …
First term, a = 3
Common difference, d = Second term – First term
⇒ 1 – 3 = -2
⇒ d = -2

(ii) -5, – 1, 3, 7 …
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, …
First term, a = 1/3
Common difference, d = Second term – First term
⇒ 5/3 – 1/3 = 4/3

(iv) 0.6, 1.7, 2.8, 3.9, …
First term, a = 0.6
Common difference, d = Second term – First term
⇒ 1.7 – 0.6 = 1.1

4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ….
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, – 6, – 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, – 4, – 8, – 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ….
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a …
(xi) a, a², a³, a⁴ …
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 1², 3², 5², 7² …
(xv) 1², 5², 7², 7³ …

Solution –

(i) 2, 4, 8, 16 …
Here, the common difference is;
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
a₄ – a₃ = 16 – 8 = 8
(a₃ – a₂) ≠ (a₂ – a₁)
So, 2, 4, 8, 16, … are not in A.P., because the common difference is not equal.

(ii) 2, 5/2 ,3, 7/2, …
a₂ – a₁ = 5/2-2 = 1/2
a₃ – a₂ = 3-5/2 = 1/2
a₄ – a₃ = 7/2-3 = 1/2
(a₄ – a₃) = (a₃ – a₂) = (a₂ – a₁)
2, 5/2 ,3, 7/2 forms an A.P. and common difference is 1/2
The next three terms are;
Fifth term a₅ = a₁ + 4d = 2 + 4 (1/2) = 4
Sixth term a₆= a₁ + 5d = 2 + 5 × 1/2 = 2 + 5/2 = (4 + 5)/2 = 9/2
Seventh term a₇= a + 6d = 2 + 6 × 1/2 = 5

(iii) – 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
(a₄ – a₃) = (a₃ – a₂) = (a₂ – a₁)
Therefore, d = -2 and the given series are in A.P.
Hence, next three terms are;
Fifth term a₅ = a_₁ + 4d = – 1.2 + 4(- 2) = -1.2 – 8 = – 9.2
Sixth term a₆ = a_₁ + 5d = – 1.2 + 5(- 2) = – 1.2 – 10 = – 11.2
Seventh term a₇ = a_₁ + 6d = – 1.2 + 6(- 2) = – 1.2 – 12 = – 13.2

(iv) – 10, – 6, – 2, 2, …
a₂ – a₁ = (-6)-(-10) = 4
a₃ – a₂ = (-2)-(-6) = 4
a₄ – a₃ = (2 -(-2) = 4
(a₄ – a₃) = (a₃ – a₂) = (a₂ – a₁)
Therefore, d = 4 and the given numbers are in A.P.
Hence, next three terms are;
Fifth Term a₅ = a_₁ + 4d = – 10 + 16 = 6
Sixth Term a₆= a_₁ + 5d = – 10 + 20 = 10
Seventh Term a₇ = a_₁ + 6d = – 10 + 24 = 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …
a₂ – a₁ = 3+√2-3 = √2
a₃ – a₂ = (3+2√2)-(3+√2) = √2
a₄ – a₃ = (3+3√2) – (3+2√2) = √2
(a₄ – a₃) = (a₃ – a₂) = (a₂ – a₁)
Therefore, d = √2 and the given series forms an A.P.
Hence, next three terms are;
Fifth term a₅ = a₁ + 4d = 3 + 4 × √2 = 3 + 4√2
Sixth term a₆ = a₁ + 5d = 3 + 5 × √2 = 3 + 5√2
Seventh term a₇ = a₁ + 6d = 3 + 6 × √2 = 3 + 6√2

(vi) 0.2, 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
(a₄ – a₃) ≠ (a₃ – a₂) ≠ (a₂ – a₁)
Therefore, and the given series doesn’t forms an A.P.

(vii) 0, -4, -8, -12 …
a₂ – a₁ = (-4)-0 = -4
a₃ – a₂ = (-8)-(-4) = -4
a₄ – a₃ = (-12)-(-8) = -4
(a₄ – a₃) = (a₃ – a₂) = (a₂ – a₁)
Therefore, d = -4 and the given series forms an A.P.
Hence, next three terms are;
Fifth term a₅ = a₁ + 4d = 0 + 4(- 4) = – 16
Sixth term a₆ = a₁ + 5d = 0 + 5(- 4) = – 20
Seventh term a₇ = a₁ + 6d = 0 + 6(- 4) = – 24

(viii) -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
(a₄ – a₃) = (a₃ – a₂) = (a₂ – a₁)
Therefore, d = 0 and the given series forms an A.P.
Hence, next three terms are;
Fifth term a₅ = a₁ + 4d = – 1/2 + 4 (0) = – 1/2
Sixth term a₆ = a₁ + 5d = – 1/2 + 5 (0) = – 1/2
Seventh term a₇ = a₁ + 6d = – 1/2 + 6 (0) = – 1/2

(ix) 1, 3, 9, 27 ….
a₂ – a₁ = 3-1 = 2
a₃ – a₂ = 9-3 = 6
a₄ – a₃ = 27-9 = 18
(a₄ – a₃) ≠ (a₃ – a₂) ≠ (a₂ – a₁)
Therefore, and the given series doesn’t forms an A.P.

(x) a, 2a, 3a, 4a …
a₂ – a₁ = 2a–a = a
a₃ – a₂ = 3a-2a = a
a₄ – a₃ = 4a-3a = a
(a₄ – a₃) = (a₃ – a₂) = (a₂ – a₁)
Therefore, d = a and the given series forms an A.P.
Hence, next three terms are;
Fifth term a₅ = a₁ + 4d = a + 4a = 5a
Sixth term a₆ = a₁ + 5d = a + 5a = 6a
Seventh term a₇ = a₁ + 6d = a + 6a = 7a

(xi) a, a², a³, a⁴ …
a₂ – a₁ = a²–a = a(a – 1)
a₃ – a₂ = a³–a²= a²(a – 1)
a₄ – a₃ = a⁴ – a³= a³(a – 1)
(a₄ – a₃) ≠ (a₃ – a₂) ≠ (a₂ – a₁)
Therefore, and the given series doesn’t forms an A.P.

(xii) √2, √8, √18, √32 …
a₂ – a₁ = √8-√2 = 2√2-√2 = √2
a₃ – a₂ = √18-√8 = 3√2-2√2 = √2
a₄ – a₃ = 4√2-3√2 = √2
(a₄ – a₃) = (a₃ – a₂) = (a₂ – a₁)
Therefore, d = √2 and the given series forms an A.P.
Hence, next three terms are;
Fifth term a₅ = a₁ + 4d = √2 + 4√2 = 5√2 = √25 × 2 = √50
Sixth term a₆ = a₁ + 5d = √2 + 5√2 = 6√2 = √36 × 2 = √72
Seventh term a₇ = a₁ + 6d = √2 + 6√2 = 7√2 = √49 × 2 = √98

(xiii) √3, √6, √9, √12 …
a₂ – a₁ = √6-√3 = √3×√2-√3 = √3(√2 – 1)
a₃ – a₂ = √9-√6 = 3-√6 = √3(√3 – √2)
a₄ – a₃ = √12 – √9 = 2√3 – √3×√3 = √3(2 – √3)
(a₄ – a₃) ≠ (a₃ – a₂) ≠ (a₂ – a₁)
Therefore, and the given series doesn’t forms an A.P.

(xiv) 1², 3², 5², 7² …
1, 9, 25, 49 …..
Here,
a₂ − a₁ = 9−1 = 8
a₃ − a₂= 25−9 = 16
a₄ − a₃ = 49−25 = 24
(a₄ – a₃) ≠ (a₃ – a₂) ≠ (a₂ – a₁)
Therefore, and the given series doesn’t forms an A.P.

(xv) 1², 5², 7², 73 …
1, 25, 49, 73 …
Here,
a₂ − a₁ = 25−1 = 24
a₃ − a₂= 49−25 = 24
a₄ − a₃ = 73−49 = 24
(a₄ – a₃) = (a₃ – a₂) = (a₂ – a₁)
Therefore, d = 24 and the given series forms an A.P.
Hence, next three terms are;
Fifth term a₅ = a₁ + 4d = 1 + 4 × 24 = 1 + 96 = 97
Sixth term a₆ = a₁ + 5d = 1 + 5 × 24 = 1 + 120 = 121
Seventh term a₇ = a₁ + 6d = 1 + 6 × 24 = 1 + 144 = 145

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NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1