**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 t_{1} =a = 15, t_{2} = 15 +8 = 23,

t_{3 = }23 +8 =31

So, the list will be as follows: 15, 23, 31, ….

Here t_{2 }– t_{1} = t_{3 }– t_{2} = …. = 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*, 3*V*/4 , (3*V*/4)^{2} , (3*V*/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

t_{2 }= a + d = 150+ 50 = 200

t_{3 = }a + 2d = 150+ 2 x 50 = 250

t_{4} = a + 3d = 150 + 150 = 300

Since t_{2} – t_{1} = t_{3} – t_{2} = 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, t_{1} = a = 10,000

t_{2}= 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

(i) a = 10, d = 10

**Answer**

Let us consider, the Arithmetic Progression series be a_{1}, a_{2}, a_{3}, a_{4}, a_{5} …

a_{1} = a = 10

a_{2} = a_{1}+d = 10+10 = 20

a_{3} = a_{2}+d = 20+10 = 30

a_{4} = a_{3}+d = 30+10 = 40

a_{5} = a_{4}+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 a_{1}, a_{2}, a_{3}, a_{4}, a_{5} …

a_{1} = a = -2

a_{2} = a_{1}+d = – 2+0 = – 2

a_{3} = a_{2}+d = – 2+0 = – 2

a_{4} = a_{3}+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 a_{1}, a_{2}, a_{3}, a_{4}, a_{5} …

a_{1} = a = 4

a_{2} = a_{1}+d = 4-3 = 1

a_{3} = a_{2}+d = 1-3 = – 2

a_{4} = a_{3}+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 a_{1}, a_{2}, a_{3}, a_{4}, a_{5} …

a_{2} = a_{1}+d = -1+1/2 = -1/2

a_{3} = a_{2}+d = -1/2+1/2 = 0

a_{4} = a_{3}+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 a_{1}, a_{2}, a_{3}, a_{4}, a_{5} …

a_{1} = a = – 1.25

a_{2} = a_{1} + d = – 1.25 – 0.25 = – 1.50

a_{3} = a_{2} + d = – 1.50 – 0.25 = – 1.75

a_{4} = a_{3} + 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

a_{2} – a_{1} = 4 – 2 = 2

a_{3} – a_{2} = 8 – 4 = 4

a_{4} – a_{3} = 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, a_{2} – a_{1} = 5/2-2 = 1/2

a_{3} – a_{2} = 3-5/2 = 1/2

a_{4} – a_{3} = 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;

a_{5} = 7/2+1/2 = 4

a_{6} = 4 +1/2 = 9/2

a_{7} = 9/2 +1/2 = 5

**(iii) -1.2, -3.2, -5.2, -7.2 …**

**Answer**

Here, a_{2} – a_{1} = (-3.2) – (-1.2) = -2

a_{3} – a_{2} = (-5.2) – (-3.2) = -2

a_{4} – a_{3} = (-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;

a_{5} = – 7.2 – 2 = -9.2

a_{6} = – 9.2 – 2 = – 11.2

a_{7} = – 11.2 – 2 = – 13.2

**(iv) -10, – 6, – 2, 2 …**

**Answer**

Here, the terms and their difference are :-

a_{2} – a_{1} = (-6) – (-10) = 4

a_{3} – a_{2} = (-2) – (-6) = 4

a_{4} – a_{3} = 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;

a_{5} = 2 + 4 = 6

a_{6} = 6 + 4 = 10

a_{7} = 10 + 4 = 14

**(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2**

**Answer**

Here, a_{2} – a_{1} = 3+√2-3 = √2

a_{3} – a_{2} = (3+2√2)-(3+√2) = √2

a_{4} – a_{3} = (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;

a_{5} = (3+√2) + √2 = 3 + 4√2

a_{6} = (3+4√2) + √2 = 3 + 5√2

a_{7} = (3+5√2) + √2 = 3 + 6√2

**(vi) 0.2, 0.22, 0.222, 0.2222 ….**

**Answer**

Here, a_{2} – a_{1} = 0.22 – 0.2 = 0.02

a_{3} – a_{2} = 0.222 – 0.22 = 0.002

a_{4} – a_{3} = 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, a_{2} – a_{1} = (-4) – 0 = -4

a_{3} – a_{2} = (-8) – (-4) = -4

a_{4} – a_{3} = (-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 :-

a_{5} = -12 – 4 = -16

a_{6} = -16 – 4 = -20

a_{7} = -20 – 4 = -24

**(viii) -1/2, -1/2, -1/2, -1/2 ….**

**Answer**

Here, a_{2} – a_{1} = (-1/2) – (-1/2) = 0

a_{3} – a_{2} = (-1/2) – (-1/2) = 0

a_{4} – a_{3} = (-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;

a_{5} = (-1/2)-0 = -1/2

a_{6} = (-1/2)-0 = -1/2

a_{7} = (-1/2)-0 = -1/2

**(ix) 1, 3, 9, 27 …**

Answer

Here,a_{2} – a_{1} = 3 – 1 = 2

a_{3} – a_{2} = 9 – 3 = 6

a_{4} – a_{3} = 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, a_{2} – a_{1} = 2a – a = a

a_{3} – a_{2} = 3a – 2a = a

a_{4} – a_{3} = 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 :-

a_{5} = 4a + a = 5a

a_{6} = 5a + a = 6a

a_{7} = 6a + a = 7a

**(xi) a, a ^{2}, a^{3}, a^{4} …**

**Answer**

Here, a_{2} – a_{1} = a^{2}–a = a(a-1)

a_{3} – a_{2} = a^{3 }–^{ }a^{2 }= a^{2}(a-1)

a_{4} – a_{3} = a^{4} – a^{3 }= a^{3}(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, a_{2} – a_{1} = √8 – √2 = 2√2 – √2 = √2

a_{3} – a_{2} = √18 – √8 = 3√2- 2√2 = √2

a_{4} – a_{3} = 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;

a_{5} = √32 + √2 = 4√2 + √2 = 5√2 = √50

a_{6} = 5√2 + √2 = 6√2 = √72

a_{7} = 6√2 + √2 = 7√2 = √98

**(xiii) √3, √6, √9, √12 …**

**Answer**

Here, a_{2} – a_{1} = √6 – √3 = √3 × √2 – √3 = √3(√2-1)

a_{3} – a_{2} = √9 – √6 = 3 – √6 = √3(√3-√2)

a_{4} – a_{3} = √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) 1 ^{2}, 3^{2}, 5^{2}, 7^{2} …**

**Answer**

Here, a_{2} − a_{1} = 9 − 1 = 8

a_{3} − a_{2 }= 25 − 9 = 16

a_{4} − a_{3} = 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) 1 ^{2}, 5^{2}, 7^{2}, 7^{3} …**

**Answer**

Here, a_{2} − a_{1} = 25−1 = 24

a_{3} − a_{2 }= 49−25 = 24

a_{4} − a_{3} = 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;

a_{5} = 73 + 24 = 97

a_{6} = 97 + 24 = 121

a_{7 }= 121 + 24 = 145

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