Exercise 5.1

NCERT Solutions for Class 10 Maths Chapter 5

Arithmetic Progressions Exercise 5.1

Page 99

 

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

Answer

(i) Let t_n be the taxi fare for first n km.
Then t₁ =a = 15, t₂ = 15 +8 = 23,
t_{3 ⁼} 23 +8 =31
So, the list will be as follows: 15, 23, 31, ….
Here t₂ – t₁ = t₃ – t₂ = …. = 8
Thus, this situation forms an Arithmetic Progression.

(ii) The amount of air present in a cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.

Answer

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 arithmetic progression.

 

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

Answer

First term a = ₹ 150
Common difference for every subsequent metre is ₹ 50
t_{1 =} a = 150
t₂ = a + d = 150+ 50 = 200
t_{3 ⁼} a + 2d = 150+ 2 x 50 = 250
t₄ = a + 3d = 150 + 150 = 300
Since t₂ – t₁ = t₃ – t₂ = 50, therefore, it is an Arithmetic Progression.

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

Answer

Let t_n be the amount of money in the nth year
Then, t₁ = a =  10,000
t₂= 10,000 + 10,000 x  8/100
= 10,000 + 800 = 10,800
t_{3 ^{=  10,800 + 10,800 x  8/100}}
= ^{_{10,800 + 864 = 11,664}}
t_{4 ^{= 11,664 + 11,664 x 8/100}}
= ^{_{11,664 + 933.12 = 12,597.12}}
The list is 10000, 10800, 11664, ^_12,597.12
Here, t2 – t1 ≠ t3 – t2
therefore, it is not an Arithmetic Progression.

 

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

(i) a = 10, d = 10

Answer

Let us consider, the Arithmetic Progression series be a₁, a₂, a₃, a₄, a₅ …
a₁ = a = 10
a₂ = a₁+d = 10+10 = 20
a₃ = a₂+d = 20+10 = 30
a₄ = a₃+d = 30+10 = 40
a₅ = a₄+d = 40+10 = 50
And so on…
∴ the Arithmetic Progression series will be 10, 20, 30, 40, 50 …
And First four terms of this Arithmetic Progression will be 10, 20, 30, and 40.

(ii) a = -2, d = 0

Answer

Let us consider, the Arithmetic Progression series be a₁, a₂, a₃, a₄, a₅ …
a₁ = a = -2
a₂ = a₁+d = – 2+0 = – 2
a₃ = a₂+d = – 2+0 = – 2
a₄ = a₃+d = – 2+0 = – 2
∴ the Arithmetic Progression series will be – 2, – 2, – 2, – 2 …
And, First four terms of this Arithmetic Progression will be – 2, – 2, – 2 and – 2.

(iii) a = 4, d = -3

Answer

Let us consider, the Arithmetic Progression series be a₁, a₂, a₃, a₄, a₅ …
a₁ = a = 4
a₂ = a₁+d = 4-3 = 1
a₃ = a₂+d = 1-3 = – 2
a₄ = a₃+d = -2-3 = – 5
∴ the Arithmetic Progression series will be 4, 1, – 2 – 5 …
And, first four terms of this Arithmetic Progression will be 4, 1, – 2 and – 5.

 

(iv) a = -1, d = 1/2

Answer

Let us consider, the Arithmetic Progression series be a₁, a₂, a₃, a₄, a₅ …
a₂ = a₁+d = -1+1/2 = -1/2
a₃ = a₂+d = -1/2+1/2 = 0
a₄ = a₃+d = 0+1/2 = 1/2
Thus, the Arithmetic Progression series will be-1, -1/2, 0, 1/2
And First four terms of this Arithmetic Progression will be -1, -1/2, 0 and 1/2.

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

Answer

Let us consider, the Arithmetic Progression series be a₁, a₂, a₃, a₄, a₅ …
a₁ = a = – 1.25
a₂ = a₁ + d = – 1.25 – 0.25 = – 1.50
a₃ = a₂ + d = – 1.50 – 0.25 = – 1.75
a₄ = a₃ + d = – 1.75 – 0.25 = – 2.00
∴ the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 ……
And first four terms of this Arithmetic Progression will be – 1.25, – 1.50, – 1.75 and – 2.00.

3. For the following Arithmetic Progressions, write the first term and the common difference.

(i) 3, 1, – 1, – 3 …

Answer

Given series,
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 …

Answer

Given series, – 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 ….

Answer

Given series, 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 …

Answer

Given series, 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 Arithmetic Progression find the common difference d and write three more terms.

(i) 2, 4, 8, 16 …

Answer

Here, the common difference is
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
a₄ – a₃ = 16 – 8 = 8
Since, a_n+1 – a_n or the common difference is not the same every time.
∴ the given series are not forming an Arithmetic Progression

 

(ii) 2, 5/2, 3, 7/2 ….

Answer

Here, a₂ – a₁ = 5/2-2 = 1/2
a₃ – a₂ = 3-5/2 = 1/2
a₄ – a₃ = 7/2-3 = 1/2
Since, a_n+1 – a_n or the common difference is same every time.
∴ d = 1/2 and the given series are in Arithmetic Progression
The next three terms are;
a₅ = 7/2+1/2 = 4
a₆ = 4 +1/2 = 9/2
a₇ = 9/2 +1/2 = 5

(iii) -1.2, -3.2, -5.2, -7.2 …

Answer

Here, a₂ – a₁ = (-3.2) – (-1.2) = -2
a₃ – a₂ = (-5.2) – (-3.2) = -2
a₄ – a₃ = (-7.2) – (-5.2) = -2
Since, a_n+1 – a_n or common difference is same every time.
∴ d = -2 and the given series are in Arithmetic Progression
Hence, next three terms are;
a₅ = – 7.2 – 2 = -9.2
a₆ = – 9.2 – 2 = – 11.2
a₇ = – 11.2 – 2 = – 13.2

(iv) -10, – 6, – 2, 2 …

Answer

Here, the terms and their difference are :-
a₂ – a₁ = (-6) – (-10) = 4
a₃ – a₂ = (-2) – (-6) = 4
a₄ – a₃ = 2 -(-2) = 4
Since, a_n+1 – a_n or the common difference is same every time.
∴ d = 4 and the given numbers are in Arithmetic Progression
Hence, next three terms are;
a₅ = 2 + 4 = 6
a₆ = 6 + 4 = 10
a₇ = 10 + 4 = 14

 

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2

Answer

Here, a₂ – a₁ = 3+√2-3 = √2
a₃ – a₂ = (3+2√2)-(3+√2) = √2
a₄ – a₃ = (3+3√2) – (3+2√2) = √2
Since, a_n+1 – a_n or the common difference is same every time.
∴ d = √2 and the given series forms a Arithmetic Progression
Hence, next three terms are;
a₅ = (3+√2) + √2 = 3 + 4√2
a₆ = (3+4√2) + √2 = 3 + 5√2

a₇ = (3+5√2) + √2 = 3 + 6√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….

Answer

Here, a₂ – a₁ = 0.22 – 0.2 = 0.02
a₃ – a₂ = 0.222 – 0.22 = 0.002
a₄ – a₃ = 0.2222 – 0.222 = 0.0002
Since, a_n+1 – a_n or the common difference is not same every time.
∴ and the given series doesn’t forms a Arithmetic Progression

(vii) 0, – 4, – 8, – 12 …

Answer

Here, a₂ – a₁ = (-4) – 0 = -4
a₃ – a₂ = (-8) – (-4) = -4
a₄ – a₃ = (-12) – (-8) = -4
Since, a_n+1 – a_n or the common difference is same every time.
∴ d = -4 and the given series forms a Arithmetic Progression
Hence, next three terms are :-
a₅ = -12 – 4 = -16
a₆ = -16 – 4 = -20
a₇ = -20 – 4 = -24

 

(viii) -1/2, -1/2, -1/2, -1/2 ….

Answer

Here, a₂ – a₁ = (-1/2) – (-1/2) = 0
a₃ – a₂ = (-1/2) – (-1/2) = 0
a₄ – a₃ = (-1/2) – (-1/2) = 0
Since, a_n+1 – a_n or the common difference is same every time.
∴ d = 0 and the given series forms a Arithmetic Progression
Hence, next three terms are;
a₅ = (-1/2)-0 = -1/2
a₆ = (-1/2)-0 = -1/2
a₇ = (-1/2)-0 = -1/2

(ix) 1, 3, 9, 27 …

Answer

Here,a₂ – a₁ = 3 – 1 = 2
a₃ – a₂ = 9 – 3 = 6
a₄ – a₃ = 27 -9 = 18
Since, a_n+1 – a_n or the common difference is not same every time.
∴ and the given series doesn’t form a Arithmetic Progression

(x) a, 2a, 3a, 4a …

Answer

Here, a₂ – a₁ = 2a – a = a
a₃ – a₂ = 3a – 2a = a
a₄ – a₃ = 4a – 3a = a
Since, a_n+1 – a_n or the common difference is same every time.
∴ d = a and the given series forms a Arithmetic Progression
Hence, next three terms are :-
a₅ = 4a + a = 5a
a₆ = 5a + a = 6a

a₇ = 6a + a = 7a

 

(xi) a, a², a³, a⁴ …

Answer

Here, a₂ – a₁ = a²–a = a(a-1)
a₃ – a₂ = a³ – a² = a²(a-1)
a₄ – a₃ = a⁴ – a³ = a³(a-1)
Since, a_n+1 – a_n or the common difference is not same every time.
∴ the given series doesn’t forms a Arithmetic Progression

(xii) √2, √8, √18, √32 …

Answer

Here, a₂ – a₁ = √8 – √2  = 2√2 – √2 = √2
a₃ – a₂ = √18 – √8 = 3√2- 2√2 = √2
a₄ – a₃ = 4√2 – 3√2 = √2
Since, a_n+1 – a_n or the common difference is same every time.
∴ d = √2 and the given series forms a Arithmetic Progression
Hence, next three terms are;
a₅ = √32 + √2 = 4√2 + √2 = 5√2 = √50
a₆ = 5√2 + √2 = 6√2 = √72
a₇ = 6√2 + √2 = 7√2 = √98

(xiii) √3, √6, √9, √12 …

Answer

Here, 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)
Since, a_n+1 – a_n or the common difference is not same every time.
∴ the given series doesn’t form a Arithmetic Progression

(xiv) 1², 3², 5², 7² …

Answer

Here, a₂ − a₁ = 9 − 1 = 8
a₃ − a₂ = 25 − 9 = 16
a₄ − a₃ = 49 − 25 = 24
Since, a_n+1 – a_n or the common difference is not same every time.
∴ the given series doesn’t form a Arithmetic Progression

(xv) 1², 5², 7², 7³ …

Answer

Here, a₂ − a₁ = 25−1 = 24
a₃ − a₂ = 49−25 = 24
a₄ − a₃ = 73−49 = 24
Since, a_n+1 – a_n or the common difference is same every time.
∴ d = 24 and the given series forms a Arithmetic Progression
Hence, next three terms are;
a₅ = 73 + 24 = 97
a₆ = 97 + 24 = 121
a₇ = 121 + 24 = 145