# KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.1

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.1.

## Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.1

Question 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 Rs. 15 for the first km and Rs. 8 for each additional km.
ii) The amount of air present in a cylinder when a vacuum pump removes the air remaining in the cylinder at a time.
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.
iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8% per annum.
Solution:
(i) Taxi fare for first km is ₹ 15
Taxi fare for second km is ₹ 15 + ₹ 8 = ₹ 23
Taxi fare for third km is ₹ 23 + ₹ 8 = ₹ 31
∴ 15, 23, 31, 39, ………………
So, This list of numbers forms an arithmetic progression with the first term a = ₹ 15
and the common difference d = ₹ 8.

ii) Let the amount of air present in a cylinder be ‘x’.

1. $$\frac{1}{4} \text { of } 1=\frac{1}{4}$$ Amount of air goes out
$$1-\frac{1}{4}=\frac{3}{4}$$ Air remaining
2. $$\frac{1}{4} \text { of } \frac{3}{4}=\frac{3}{16}$$ Amount of air goes out
$$\frac{3}{4}-\frac{3}{16}=\frac{9}{16}$$ Air remaining
3. $$\frac{1}{4} \text { of } \frac{9}{16}=\frac{9}{64}$$ Amount of air goes out
$$\frac{9}{16}-\frac{9}{64}=\frac{27}{64}$$ Air remaining

Set $$\frac{3}{4}, \frac{9}{16}, \frac{27}{64}, \ldots .$$
$$\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=\frac{9}{16}-\frac{3}{4} \quad \mathrm{d}=\frac{-3}{16}$$
$$\mathrm{d}=\mathrm{a}_{3}-\mathrm{a}_{2}=\frac{27}{64}-\frac{9}{16} \quad \mathrm{d}=\frac{-9}{64}$$
Here the value of ‘d’ is not constant.
∴ The given list of numbers does not form an A.P.

(iii) Cost of Digging for the

d = a2 – a1 = 200 – 150 = 50
d = a3 – a2 = 250 – 200 = 50
Here value of ‘d’ is constant.
This forms an AP.

iv) Amount of money deposited in the
d = a2 – a1 = 11,664 – 10.800 = Rs. 864
d = a3 – a2 = 12,597,12 – 11.664 = Rs. 933.12
Here the value of ‘d’ is not constant.
∴ The given list of numbers does not form an AP.

Question 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 = 1/2
(y) a = -1.25, d = -0.25
Solution:
(i) ∵ Tn = a + (n – 1)d
∴ For a = 10 and d = 10, we have:
T1 =10 + (1 – 1) × 10 = 10 + 0 = 10
T2 = 10 + (2 – 1) × 10 = 10 + 10 – 20
T3 = 10 + ( 3 – 1) × 10 = 10 + 20 = 30
T4 = 10 + (4 – 1) × 10 = 10 + 30 = 40
Thus, the first four terms are:
10, 20, 30, 40

(ii) ∵ Tn = a + (n – 1)d
∴ For a = -2 and d = 0,we have:
T1 = -2 + (1 – 1) × 0 = -2 + 0 = -2
T2 = -2 + (2 – 1) × 0 = -2 + 0 = -2
T3 = -2 + (3 – 1) × 0= -2 + 0 = -2
T4 = -2 + (4 – 1) × 0 = -2 + 0 = -2
∴ Thus, the first four terms are: -2, -2, -2, -2

(iii) ∵ Tn = a + (n – 1)d
For a = 4 and d = -3, we have:
T1 = 4 + (1 – 1) × (-3) = 4 + 0 = 4
T2 = 4 + (2 – 1) × (-3) = 4 + (-3) = 1
T3 = 4 + (3 – 1) × (-3) = 4 + (-6) = -2
T4 = 4 + (4 – 1) × (-3) = 4 + (-9) = -5
Thus, the first four terms are:
4, 1, -2, -5

(iv)

(v) ∵ Tn = a + (n – 1)d
∴ For a = -1.25 and d = -0.25, we have
T1 = -1.25 + (1 – 1) × (-0.25) = -1.25 + 0
= -1.25
T2 = -1.25 + (2 – 1) × (-0.25) = -1.25 + (-0.25) = -1.50
T3 = -1.25 + (3 – 1) × (-0.25) = -1.25 + (-0.50) = -1.75
T4 = -1.25 + (4 – 1) × (-0.25) = -1.25 + (-0.75) = -2.0

Question 3.
For the following APs, write the first term and the common difference :
(i) 3, 1,-1,-3, ………..
(ii) -5, -1, , 7, …………
(iii) $$\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \ldots \ldots$$
(iv) 0.6, 1.7, 2.8, 3.9, ………..
Solution:
i) 3, 1, – 1, – 3, ….
∴First term = a = 3
Common difference d = a2 – a1
= 1 – 3 = -2.

ii) -5, -1, 3, 7
∴ First term = a = – 5
Common difference = d = a2 – a,
= – 1 – (- 5) = -1 +5 = 4

iii) $$\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}$$ , ………..
∴First term = a = $$\frac{1}{3}$$
Common difference = d = a2 – a1
= $$\frac{5}{3}-\frac{1}{3}=\frac{4}{3}$$

iv) 0.6, 1.7, 2.8, 3.9,….
∴ First term = a = 0.6
Common difference = d = a2 – a1
= 1.7 – 0.6 = 1.1

Question 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) $$2, \frac{5}{2}, 3, \frac{7}{2}, \ldots \ldots$$
iii) -1.2, -3.2, -5.2, -7.2
iv) -10, -6, -2, 2, …….
v) $$3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2} \ldots \ldots$$
vi) 0.2, 0.22, 0.222, 0.2222, ………
vii) 0, -4, -8, -12, ………
viii) $$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots \ldots$$
ix) 1, 3, 9, 27, ……..
x) a, 2a, 3a, 4a, ………..
xi) a, a2, a3, a4, ………..
xii) $$\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32} \ldots$$
xiii) $$\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}$$
xiv) 11, 32, 52, 72,
xv) 11, 52, 72, 73, ……….
Solution:
i) 2, 4, 8, 16, ………
a2 – a1 = 4 – 2 = 2
a3 – a2 = 8 – 4 = 4
a2 – a1 ± a3 – a2
∴The given list of numbers does not form an A.P

ii) $$2, \frac{5}{2}, 3, \frac{7}{2}, \ldots \ldots$$
$$\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=\frac{5}{2}-\frac{2}{1}=\frac{5-4}{2}=\frac{1}{2}$$
$$\mathrm{d}=\mathrm{a}_{3}-\mathrm{a}_{2}=\frac{3}{1}-\frac{5}{2}=\frac{6-5}{2}=\frac{1}{2}$$
Here ‘d’ is constant.
∴ the given List of numbers form an AP.
Next three terms are: $$4,\frac{9}{2}, 5$$

(iii) – 1.2, – 3.2, – 5.2, – 7.2
a2 – a1 =- 3.2 – (- 1.2) = – 3.2 + 1.2 = – 2
a3 – a2 = – 5.2 – (- 3.2) = – 5.2 + 3.2 = – 2
a4 – a3=- 7.2 – (- 5.2) – 7.2 + 5.2 = – 2
a2 – a1 = a3 – a2 = a4 – a3
∴The given list of numbers form an A.P Common difference = d = – 2
∴Next three terms are – 7.2 + (- 2),
– 7.2 + 2(- 2), – 7.2 + 3(- 2)
∴ – 9.2, – 11.2, – 13.2

iv) – 10, – 6, – 2, 2, ….
a2 – a1 = – 6- (- 10) = – 6 + 10 = 4
a4– a3 – 2 – (- 2) 2 +24
∴a2 – a3 = a3 – a2= a4 – a3
∴The given list of numbers form an A.P
Common difference = d =4
Next three terms 2 + 4, 2 + 2(4), 2 + 3(4)
∴ 6, 10, 14

v) $$3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2} \ldots \ldots$$
d = a2 – a1 = $$=3+\sqrt{2}-3$$
$$\mathrm{d}=\sqrt{2}$$
d = a3 – a2 = $$=3+2 \sqrt{2}-(3+\sqrt{2})$$
$$=3+2 \sqrt{2}-3+\sqrt{2}$$
$$\mathrm{d}=\sqrt{2}$$
Here ‘d’ is constant
∴ Given set of numbers forms an A.P.
Further three terms are:
$$3+3 \sqrt{2}+\sqrt{2}=3+4 \sqrt{2}$$
$$3+4 \sqrt{2}+\sqrt{2}=3+5 \sqrt{2}$$
$$3+5 \sqrt{2}+\sqrt{2}=3+6 \sqrt{2}$$

vi) 0.2, 0.22, 0.222, 0.2222,…
a2 – a1 =0.22- 0.2 = 0.02
a31, = 0.222 – 0.22 = 0.002
a4 – a3 = 0.2222 – 0.222 = 0.0002
a4– a3 = a3– a2 = a2– a1
∴ The given list of numbers does not form an A.P.

vii) 0, – 4, – 8, – 12,…
a2 – a1 – 4 – (0) = – 4
a3 – a2 =- 8 – (- 4) = – 8 +4 = – 4
a4 – a3 =- 12- (- 8) = – 12 + 8 = – 4
a2 – a1 = a3 – a2 = a4 – a3
∴ The given list of numbers form an A.P
Common difference d = – 4
Next three terms are
– 12 + (- 4), – 12 + 2(- 4), – 12 + 3(- 4)
– 12 – 4, – 12 – 8, – 12 – 12
∴ -16, – 20, – 24

viii) $$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots \ldots$$
$$a_{1}=-\frac{1}{2}, a_{2}=-\frac{1}{2}$$
$$\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=-\frac{1}{2}-\left(-\frac{1}{2}\right)$$
$$d=-\frac{1}{2}+\frac{1}{2}=0$$
$$\mathrm{d}=\mathrm{a}_{3}-\mathrm{a}_{2}=-\frac{1}{2}-\left(-\frac{1}{2}\right)$$
$$\mathrm{d}=-\frac{1}{2}+\frac{1}{2}=0$$
Here d is constant.
∴ Given set of numbers form an AP.
Next three terms are:
$$-\frac{1}{2}+0=-\frac{1}{2}$$
$$-\frac{1}{2}+0=-\frac{1}{2}$$
$$-\frac{1}{2}+0=-\frac{1}{2}$$
∴ Next three terms are: $$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}$$

ix) 1, 3, 9, 27,…
a2 – a1 = 3 – 12
a3 – a2 = 9 – 3 = 6
a4 – a3 – 27 – 9 – 18
∴a2 – a1 ≠ a3 – a2 = a4 – a3
The given list of numbers does not form an A.P.

x) a, 2a, 3a, 4a,…
a2 – a1 = 2a – a = a
a3 – a2 = 3a – 2a = a
a4 – a3 = 4a -3a = a
∴ a2 – a1 = a3 – a2 = a4 – a3
Hence, the given list of numbers form anA.P
∴Common difference = d = a.
∴Next three terms 4a + a, 4a + 2(a), 4a + 3(a)
∴5a, 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)
a2 – a1 ≠ a3 – a2 = a4 – a3
∴ The given list of numbers does not form an A.P.

xii) $$\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32} \ldots$$
$$a_{1}=\sqrt{2}, a_{2}=\sqrt{8}$$
$$\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=\sqrt{8}-\sqrt{2}$$
$$\mathrm{d}=2 \sqrt{2}-\sqrt{2}=\sqrt{2}$$
$$\mathrm{d}=\mathrm{a}_{3}-\mathrm{a}_{2}=\sqrt{18}-\sqrt{8}$$
$$\mathrm{d}=3 \sqrt{2}-2 \sqrt{2}=\sqrt{2}$$
Here d’ is constant.
∴ Given a set of numbers form an A.P.
Succeding three terms are :
$$\sqrt{32}+\sqrt{2}=4 \sqrt{2}+\sqrt{2}=5 \sqrt{2}$$
$$5 \sqrt{2}+\sqrt{2}=6 \sqrt{2}$$
$$6 \sqrt{2}+\sqrt{2}=7 \sqrt{2}$$
$$5 \sqrt{2}, 6 \sqrt{2}, 7 \sqrt{2}$$
∴ Next three terms are: $$\sqrt{50}, \sqrt{72}, \sqrt{98}$$

xiii) $$\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}$$
$$a_{1}=\sqrt{3}, a_{2}=\sqrt{6}$$
$$\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=\sqrt{6}-\sqrt{3}$$
$$\mathrm{d}=\mathrm{a}_{3}-\mathrm{a}_{2}=\sqrt{9}-\sqrt{6}$$
$$\mathrm{d}=3-\sqrt{6}$$
Here ‘d is not constant.
∴ Given a set of numbers does not form an A.P.

xiv) 11,32,52, 72, …………
1, 9, 25, 49, ………….
a1 = 1, a2 = 9
d = a2 – a1 = 9 – 1
d = 8
d = a3 – a2 = 25 – 9
d = 16
Here ‘d’ is not constant,
∴ Given set of numbers do not form an A.P.

xv) 12, 52, 72, 72, …….
a2 – a1 = 12 = 25 – 1 = 24
a3 – a2 72 – 52 = 49 – 25 = 24
a4 – a3 = 73 – 49 = 24
∴ The given list of numbers forms an A.P
Common difference = d = 24
Next three terms are 73 + 24, 73 + 2 × 24, 73 + 3 × 24
∴ 97, 73 + 48, 73 + 72
∴ 97, 121, 145

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.1, drop a comment below and we will get back to you at the earliest.

### 1 thought on “KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.1”

1. its very useful and to solve the sum easily
THANK YOU