**NCERT Solution For Class 10, Maths, Chapter 5 Arithmetic Progressions, Exercise 5.1** has questions related basics of the this chapter. With the help of suitable formula, students can easily solve all the problems. Solutions of class 10 , maths chapter 5 , exercise 5.1 is given below.

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Toggle## Class 10, Maths, Chapter 5, Exercise 5.1 Solutions

**Q.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.**

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

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

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

**Ans:**

**(i) According to the statement,** the fare

1 km = Rs 15

2 km = Rs (15+8) = 23

3 km = Rs [15+(2×8)] = 31

4 km = Rs [15 + (3 x 8)] = 39

Now, we have series, 15, 23, 31, 39 ……….

Calculation of common difference,

d = 23 – 15 = 8

d = 31 – 23 = 8

d = 39 – 31 = 8

Since the common difference is same. So, **it is an AP**

**(ii) Let the amount of air present in the cylinder 64 liter. **Now, each time ¼^{th} air of total air removed by vacuum pump. Therefore,

64 liter

= 64 – $\left( \frac{1}{4}\times 64 \right)$ = 64 – 16 = 48 litre

= 48 – $\left( \frac{1}{4}\times 48 \right)$ = 48 – 12 = 36 litre

= 36 – $\left( \frac{1}{4}\times 36 \right)$ = 36 – 9 = 27 litre

Now, we have series,64, 48, 36, 27 ……….

Calculation of common difference,

d = 48 – 64 = -16

d = 36 – 48 = -12

d = 27 – 36 = – 9

Since the common difference is not same. So, it not AP.

**(iii) According to the statement,** the cost of digging in Rs for –

1 m = 150

2 m = (150 + 50) = 200

3 m = [150 + (2 x 50)] = 250

4 m = [150 + (3 x 50)] = 300

Now, we have series, 150, 200, 250, 300 ……….

Calculation of common difference,

d = 200 – 150 = 50

d = 250 – 200 = 50

d = 300 – 250 = 50

Since the common difference is same. So, **it is an AP**

**Q.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 **

**(ii) a = –2, d = 0 **

**(iii) a = 4, d = – 3 **

**(iv) a = – 1, d = $\frac{1}{2}$ **

**(v) a = – 1.25, d = – 0.25**

** ****Ans: **We know that if the first term is a and the common difference is d, then

a, a + d, a i- 2d, a + 3d, … represents an AP for different values of a and d

**(i) Putting a = 10, d = 10, we get**

a_{n} = a + (n-1) d

a_{1} = 10 + (1-1) 10 = 10

a_{2} = 10 + (2-1) 10 = 20

a_{3} = 10 + (3-1) 10 = 30

a_{4} = 10 + (4-1) 10 = 40

So, the required AP = 10, 20, 30, 40………

**(ii) Putting a = 2, d = 0**

a_{n} = a + (n-1) d

a_{1} = – 2 + (1-1) 0 = – 2

a_{2} = – 2 + (2-1) 0 = -2

a_{3} = – 2 + (3-1) 0 = -2

a_{4} = – 2 + (4-1) 0 = -2

So, the required AP = -2, -2, -2, -2, ………

**(iii) Putting a = 4, d = – 3**

a_{n} = a + (n-1) d

a_{1} = 4 + (1-1) -3 = 4

a_{2} = 4 + (2-1) -3 = 1

a_{3} = 4 + (3-1) -3 = -2

a_{4} = 4 + (4-1) -3 = -5

So, the required AP = 4, 1, -2, -5 ………

**(v) Putting a = -1.25, d = – 0.25**

a_{n} = a + (n-1) d

a_{1} = -1.25 + (1-1) – 0.25 = -1.25

a_{2} = -1.25 + (2-1) – 0.25 = -1.50

a_{3} = -1.25 + (3-1) – 0.25 = -1.75

a_{4} = -1.25 + (4-1) – 0.25 = – 2.0

So, the required AP = -1.25, -1.50, -1.75, -2.0 ………

**Q.3. For the following APs, write the first term and the common difference:**

**(i) 3, 1, – 1, – 3, . . . **

**(ii) – 5, – 1, 3, 7, . . . **

**(iii) $\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3}…..$ **

**(iv) (iv) 0.6, 1.7, 2.8, 3.9, . . .**

**Ans: ****(i) The given AP is 3, 1,-1,- 3, …**

** **First term, a = 3

Common difference, d = a_{2} – a_{1} = 1 – 3 = -2

Clearly, a = 3 and d = – 2.

**(ii) The given AP is- 5,-1, 3, 7, …**

First term, a = -5

Common difference, d = a_{2} – a_{1} = -1 – (-5) = 4

Clearly, a =- 5 and d = 4

**(iii) The given AP is $\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3}…..$**

First term, a = ** $\frac{1}{3}$**

Common difference, d = a_{2} – a_{1} =**$\frac{5}{3}-\frac{1}{3}=\frac{5-1}{3}=\frac{4}{3}$**

Clearly, a = ** $\frac{1}{3}$**and d = **$\frac{4}{3}$ **

**(iv) The given AP is 0.6, 1.7, 2.8, 3.9, …**

First term, a = 0.6

Common difference, d = a_{2} – a_{1} = 1.7 – 0.6 = 1.1

Clearly, a = 0.6 and d = 1.1

**Q.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, . . . **

**(ii) $2,\frac{5}{2},3,\frac{7}{2},….$ **

**(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . **

**(iv) -10, -6, -2, 2,… **

**(v) $3,\,\,3+\sqrt{2},\,\,3+2\sqrt{2},\,\,3+3\sqrt{2},……$ **

**(vi) 0.2, 0.22, 0.222, 0.2222, . . .**

**(vii) 0, – 4, – 8, –12, . . . **

**(viii) $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},….$**

**(ix) 1, 3, 9, 27, . . . **

**(x) a, 2a, 3a, 4a, . . .**

**(xi) a, a ^{2}, a^{3}, a^{4}, . . . **

**(xii) $\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},$**

**(xiii) $\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},$ **

**(xiv) 1 ^{2}, 3^{2}, 5^{2}, 7^{2}, . . .**

**(xv) 1 ^{2}, 5^{2}, 7^{2}, 73, . . .**

**Ans:**

(i) **2, 4, 8, 16, . . .**

Common difference, d =

Here, a_{2} – a_{1} = 4 -2 = 2

a_{3} – a_{2} = 8 – 4 = 4

And, a_{4} – a_{3} = 16 – 8 = 8

So, the difference is not same. Thus, 2, 4, 8, 16, ….. does not form an AP.