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

KSEEB Solutions For Class 10 Maths 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 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 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) Rent for            Rent for            Rent for
1st km                  2nd km             3rd km
Rs. 15                     Rs. 8                 Rs. 8
d = a2 – a1 = 8 – 15 = -7
d = a3 – a2 = 8 – 8 = 0
Here ’d’ is not constant.
Hence it does not form an AP.

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 value of ‘d’ is not constant.
∴ The given list of numbers does not form an AP.

KSEEB Solutions For Class 10 Maths Arithmetic Progression 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 = $$\frac{1}{2}$$
v) a = -1.25, d = 0.25
Solution:
i) a = 10, d = 10
First four terms are,
a, a + d, a + 2d, a + 3d
10, 10 + 10, 10 + 20, 10 + 30
10. 20, 30, 40.

ii) a = -2, d = 0
First four terms are,
a, a + d, a + 2d, a + 3d
-2. -2 + 0, -2 + 0, -2 + 0
-2. -2. -2, -2

iii) a = 4, d = -3
First four terms are,
a, a + d, a + 2d, a + 3d
4, 4 – 3, 4 – 2 × 3, 4 – 3 ×3
4, 1, -2, -5

iv) a = -1, d = $$\frac{1}{2}$$
First four terms are,
a, a + d, a + 2d, a + 3d
$$-1,-1+\frac{1}{2},-1+2 \times \frac{1}{2},-1+3 \times \frac{1}{2}$$
$$-1,-\frac{1}{2}, 0, \frac{1}{2}$$

v) a = -1.25, d = 0.25
First four terms are,
a, a + d, a + 2d, a + 3d
-1.25, -1.25 + 0.25, -1.25 + 2 × 0.25, -1.25 + 3 ×0.25
-1.25, -1, -0.75, -0.50

Arithmetic Progression Exercise 1.1 KSEEB Solutions 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
d = -2.

(ii) -5, -1, 3, 7, ……….
First term, a = -5,
Common difference, d = a2 – a1 = -1 – (-5)
d = -1 + 5
d = 4.

(iii) $$\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \ldots \ldots$$
First term, $$\mathrm{a}=\frac{1}{2}$$
Common difference, d = a2 – a1
$$=\frac{1}{2}-\frac{1}{2}$$
d= 0

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

Arithmetic Progression Class 10 Exercise 1.1 Solutions 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…………
d = a2 – a1= 4 – 2 = 2
d = a2 – a1 = 8 – 4 = 4
Here ‘d’ is not constant.
∴ the given List of numbers does not form an AP.

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
d = a2 – a1 = -3.2 – (-1.2) = -3.2+ 1.2
d = -2
d= a3 – a2 = -5.2 – (-3.2) = -5.2 + 3.2
d = -2
Here d is constant.
∴ This is an Arithmetic Progression.
Next three terms are:
-7.2 – 2 = -9.2
-9.2 – 2 = -11.2
-11.2 – 2 = -13.2
-11.2-2 = -13.2

iv) -10. -6. -2, 2,..,…
d = a2 – a1 = -6 – (- 10) = -6 + 10
d = 4
d = a3 – a2= -2 – (-6) = -2 + 6
d = 4
Here d is constant.
∴ Given set of numbers form an A.P.
Further three terms are
2 + 4 = 6
4 + 4 = 8
8 + 4 = 12

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
d = a2 – a1 = 0.22 – 0.2
d = 0.02
d = a3 – a2 = 0.222 – 0.22
d = 0.002
Here d is not constant,
∴ Given set of numbers do not form an A.P.

vii) 0,-4, -8, -12, …………
a1 = 0, a2 = -4
d = a2 – a1 = -4 – 0
d = -4
d = a3 – a2 = -8 – (-4) = -8 + 4
d = -4
Here d is constant.
∴ Given set of numbers form an AP.
Further three terms are:
-12 – 4 = -16
-16 – 4 = -20
-20 – 4 = -24
∴ Next three terms are: -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
d = a2 – a1 = 3—1
d = 2
d = a3 – a2
d = 6
Here ‘d’ not is constant
∴ Given set of numbers do not form an A.P.

x) a, 2a, 3a,4a
a1= a a2 = 2a
d = a2 – a1 = 2a-a
d = a
d = a3 – a2 = 3a-2a
d = a
Here d’ is constant.
∴ Given set of numbers form an A.P.
Further three terms are :
4a + a = 5a
5a + a = 6a
6a + a = 7a
∴ Next three terms are: 5a, 6a, 7a.

xi) a, a2, a3, a4
a1 = a. a2 = a2
d = a2 – a1 = a2 – a
d= a(a – 1)
d = a3 – a2 = a3 – a2
d = a2 (a – 1)
Here ‘d is not constant.
∴ Given set of numbers do 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 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 set of numbers do 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) 11, 52, 72, 73, ……….
1, 25. 49, 73, ……….
a1 = 1. a2 = 25, a3 = 49, a4 = 73
d = a2 – a1 = 25 – 1
d = 24
d = a3 – a2 = 49 – 25
d = 24
d = a4 – a3 = 73 – 49
d = 24
Here ‘d’ is constant.
∴ Given set of numbers form an A.P.
Succeeding three terms are:
73 + 24 = 97
97 + 24 = 121
121 + 24 = 145
∴ Next three terms are: 97, 121, 145.

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