KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1

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Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1

Question 1.
In which of the following situation, 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) 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) The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder at a time.
Answer:
The amount of air present in the cylinder = a1 = x units.
Amount of air present in the cylinder after the first-time removal
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 5

Amount of air present in the cylinder after second-time removal
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 1

a2 – a1 = a3 – a2
∴ a, a2, a3 does not form an A.P

(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.
Answer:
Cost of digging the well for first metre = a1 = ₹ 150
Cost of digging the well for second metre = a2 = 150 + 50 = ₹ 200
Cost of digging the well for third metre = a3 = 200 + 50 = ₹ 250
∴ a1, a2, a3,…….
150, 200, 250, ………
a2 – a1 = a3 – a2 = a4 – a3
200 -150 = 250 – 200 = 300 – 250
50 = 50 = 50
So this list of numbers form an A.P. therefore the common difference, d = ₹ 50.

(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum.
Answer:
We know that amount of present value p at r% compound interest after ‘n’ years.

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 2
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 2

a2 – a1 ≠ a3 – a2
∴ a1, a2, a3 does not from an A.P

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
Answer:
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 = \(\frac{1}{2}\)
Four terms of an A.P are a, a + d, a + 2d, a + 3d.
First term = a = – 1
Second term = a + d
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 6

∴ Four terips of an AP are
\(-1, \frac{-1}{2}, 0, \frac{1}{2}\)

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

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1

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) 0.6, 1.7, 2.8, 3.9,…
Answer:
(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

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) KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 7
(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, a2, a3, a4,…
(xii) KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 9
(xiii) KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 8
(xiv) 12, 32,52, 72,………..
(xv) 12, 52, 72, 73,…….
Answer:
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) \(2, \frac{5}{2}, 3, \frac{7}{2}\),………..

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 10
∴ a2 – a1 = a3 – a2 = a4 – a3
∴ The given list of numbers from an A.P common difference = \(\frac{1}{2}\)
∴ Next three terms

∴ \(4, \frac{9}{2}, 5\)

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

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1

v)
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 7
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 8

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 a 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.

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 9
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 10

Hence the given list of numbers form an A.P
Common difference = d = O
∴ Next three terms are
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 11

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 a 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.

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 12
∴ a – a1 ≠ a3 – a2 = a1 – a3
∴The given list of numbers form an A.P

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 13

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1 14

a2 – a1 ≠ a3 – a2
∴ The given list of numbers does not form anA.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.

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.1

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