AP 10th Class Maths 5th Chapter Arithmetic Progressions Exercise 5.1 Solutions

Well-designed AP 10th Class Maths Textbook Solutions Chapter 5 Arithmetic Progressions Exercise 5.1 offers step-by-step explanations to help students understand problem-solving strategies.

Arithmetic Progressions Class 10 Exercise 5.1 Solutions – 10th Class Maths 5.1 Exercise Solutions

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 ₹ 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 \(\frac{1}{4}\) 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.
Solution:
i) Yes.
Fare for first km = ₹ 15
Fare for next 1 km = ₹ 15 + ₹ 8 = ₹ 23
Fare for next 2 km = 15 + 8 + 8
= ₹ 31 ………. etc.
So, 15,23,31,…….. forms an Arithmetic Progression as each succeeding term is obtained by adding 8 its preceeding term.

ii) No.
Volume of air in present = V
If \(\frac{1}{4}\) of air removes from remaining = \(\frac{3}{4}\) V.
If \(\frac{1}{4}\) of air removes from remaining
= \(\frac{3}{4}\) of \(\frac{3}{4}\)V = \(\left(\frac{3}{4}\right)^2\)V ……… etc.
So, \(\frac{3}{4}\)V, \(\left(\frac{3}{4}\right)^2\)V, it is not in the form of Arithmetic Progression. Because, each succeeding term is not obtained by addition of its preceeding term.

iii) Yes.
Cost of digging a well for first metre = ₹ 150
Cost of digging for next one metre = ₹ 150 + ₹ 50 = ₹ 200
Cost of digging for next two metres = ₹ 150 + ₹ 50 + ₹ 50 = ₹ 250 …….. etc.
So, 150, 200,250, forms an Arithmetic Progression as each succeeding term is obtained by adding 50 its preceeding term.

iv) No.
Amount in the first year = 10000(1 + \(\frac{8}{100}\))
Amount at the end of the second year = 10000(1 + \(\frac{8}{100}\))²
Amount at the end of the third year
= 10000\(\left(1+\frac{8}{100}\right)^3\) ……. etc.
So, 10000(1 + \(\frac{8}{100}\)), 10000(1 + \(\frac{8}{100}\))²,
10000(1 + \(\frac{8}{100}\))³, ……….. cannot find an Arithmetic Progression.
Because, each succeeding term is not obtained by addition of its pre-ceeding term.

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) Given first term a = 10,
Common difference (d) = 10
The arithmetic progression with first term a and common difference d is given by
a, a + d, a + 2d, a + 3d, a + 4d, …..
10, 10 + 10, 10 + 2(10), 10 + 3(10),
10 + 4(10),
10, 20, 30, 40, 50, …..
Therefore, AP is 10, 20, 30, 40, 50,

ii) Given first term a = -2,
Common difference d = 0
The arithmetic progression with first term a and common difference d is given by
a, a + d, a + 2d, a + 3d, a + 4d,
-2, – 2 + 0, – 2 + 2(0), -2 + 3(0),
– 2 + 4(0),…..
-2, -2, -2, -2, -2,
Therefore, AP is -2, -2, -2, -2, -2,

iii) Given first term a = 4,
Common difference d = -3
The arithmetic progression with first term a and common difference d is given by
a, a + d, a + 2d, a + 3d, a + 4d,
4, 4 + (-3), 4 + 2(-3), 4 + 3(-3), 4 + 4(-3),
4, 4 – 3, 4 – 6, 4 – 9, 4 – 12, ………..
4, 1, -2, -5, -8, ……….
Therefore, AP is 4, 1, -2, -5, -8,

iv) Given first term a = -1,
Common difference d = \(\frac{1}{2}\)
The arithmetic progression with first term a and common difference d is given by
a, a + d, a + 2d, a + 3d, a + 4d,

v) Given first term a = -1.25,
Common difference d = – 0.25
The arithmetic progression with first term a and common difference d is given by
a, a + d, a + 2d, a + 3d, a + 4d,
-1.25,-1.25 + (-0.25), -1.25 + 2(-0.25), -1.25 + 3(-0.25), ……….
-1.25, -1.25 – 0.25, -1.25 – 0.50, -1.25,-0.75 ……….
-1.25, -1.50, -1.75, -2.00 ………..
Therefore, AP is -1.25, -1.50, -1.75, -2.00 ……….

Question 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,…………..
Solution:
i) Given AP is 3, 1, -1, -3,.. .
First term a = 3,
Common difference d = a₂ – a₁ = 1 – 3
∴ d = -2

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

iii) Given AP is \(\frac{1}{3}\), \(\frac{5}{3}\), \(\frac{9}{3}\), \(\frac{13}{3}\),……….
First term a = \(\frac{1}{3}\)
Common difference d = a₂ – a₁
= \(\frac{5}{3}\) – \(\frac{1}{3}\) = \(\frac{4}{3}\)

iv) Given AP is 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

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, \(\frac{7}{2}\),…..
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, a², a³, a⁴,…….
xii) \(\sqrt{2}\), \(\sqrt{8}\), \(\sqrt{18}\), \(\sqrt{32}\),…
xiii) \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{9}\), \(\sqrt{12}\),……
xiv) 1², 3², 5², 7²,……..
xv) 1², 5², 7², 73,……..
Solution:
i) Given, 2, 4, 8, 16,……..
We have common difference,
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
a₄ – a₃ = 16 – 8 = 8
That is common difference is not same every time. So, given list of numbers does not form an A.P.

ii) Given, 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\),……….
We have common difference,
a₂ – a₁ = \(\frac{5}{2}\) – 2 = \(\frac{1}{2}\)
a₃ – a₂ = 3 – \(\frac{5}{2}\) = \(\frac{1}{2}\)
a₄ – a₃ = \(\frac{7}{2}\) – 3 = \(\frac{1}{2}\)
That is common difference a₂ – a₁ = a₃ – a₂ = a₄ – a₃ …… is the same every time.
So, the given list of numbers forms an A.P. with the common difference (d) = \(\frac{1}{2}\).
∴ Next three terms are 4, \(\frac{9}{2}\), 5, ……….

iii) Given, -1.2, – 3.2, -5.2, – 7.2,……
We have common difference,
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
That is common difference a₂ – a₁ = a₃ – a₄ = a₄ – a₃ ……… is the same every time.
So, the given list of numbers forms an A.P with the common difference d = -2.
Next three terms are -9.2, -11.2, -13.2,……..

iv) Given -10, -6, -2, 2,
We have common difference
a₂ – a₁ = -6 – (-10) = -6 + 10 = 4
a₃ – a₂ = -2 – (-6) = -2 + 6 = 4
a₄ – a₃ = 2 – (-2) = 2 + 2 = 4
That is common difference a₂ – a₁ = a₂ – a₂ = a₄ – a₃…….. is the same every time.
So, the given list of numbers form an A.P with the common difference d = 4.
∴ Next three terms are 6, 10, 14.

v) Given, 3, 3 + \(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\) ,… We have Common difference,
a₂ – a₁ = 3 + \(\sqrt{2}\) – 3 = \(\sqrt{2}\)
a₃ – a₂ = 3 + 2\(\sqrt{2}\) – (3 + \(\sqrt{2}\))
= 3 + 2\(\sqrt{2}\) – 3 – \(\sqrt{2}\) = \(\sqrt{2}\)
a₄ – a₃ = 3 + 3\(\sqrt{2}\) – (3 + 2\(\sqrt{2}\))
= 3 + 3\(\sqrt{2}\) – 3 – 2\(\sqrt{2}\) = \(\sqrt{2}\)
That is common difference a₂ – a₁ = a₃ – a₂ = a₄ – a₃…… is the same every time.
So, the given list of numbers forms an A.P with the common difference d = \(\sqrt{2}\).
∴ Next three terms are 3 + 4\(\sqrt{2}\), 3 + 5\(\sqrt{2}\), 3 + 6\(\sqrt{2}\),……..

vi) Given 0.2, 0.22, 0.222, 0.2222,………
We have common difference,
a₂ – a₁ = 0.22 – 0.2 = 0.02
a₃ – a₂ = 0.222 – 0.22 = 0.002
a₄ – a₃ = 0.2222 – 0.222 = 0.0002
That is common difference a₂ – a₂ = a₃ – a₂ ≠ a₄ – a₃ … is not same every time.
So, the given list of numbers does not form an A.P.

vii) Given 0, – 4, – 8, -12,……
We have common difference,
a₂ – a₁ = -4 – 0 = -4
a₃ – a₂ = -8 – (-4) = – 4
a₄ – a₃ = -12 – (-8) = -4
That is common difference a₂ – a₁ = a₃ – a₂ = a₄ – a₃ ……… is the same every time.
So, the given list of numbers forms an A.P with the common difference d = -4.
∴ Next three terms are -16, -20, -24,………

viii) Given
We have common difference,
a₂ – a₁ = –\(\frac{1}{2}\) – (-\(\frac{1}{2}\)) = –\(\frac{1}{2}\) + \(\frac{1}{2}\) = 0
a₃ – a₂ = –\(\frac{1}{2}\) – (-\(\frac{1}{2}\)) = –\(\frac{1}{2}\) + \(\frac{1}{2}\) = 0
a₄ – a₃ = –\(\frac{1}{2}\) – (-\(\frac{1}{2}\)) = –\(\frac{1}{2}\) + \(\frac{1}{2}\) = 0
That is common difference a₂ – a₁ = a₃ – a₂ = a₄ – a₃ … is the same every time.
So, the given list of numbers forms an A.P with the common difference d = 0.
∴ Next three terms are –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\),………..

ix) Given 1, 3, 9, 27,…….
We have common difference,
a₂ – a₁ = 3 – 1 = 2
a₃ – a₂ = 9 – 3 = 6
a₄ – a₃ = 27 – 9 = 18
That is common difference a₂ – a₁ ≠ a₃ – a₂ ≠ a₄ – a₃ ……… is not same every time.
So, the given list of numbers does not form an A.P.

x) Given a, 2a, 3a, 4a,……….
We have common difference,
a₂ – a₁ = 2a – a = a
a₃ – a₂ = 3a – 2a = a
a₄ – a₃ = 4a – 3a = a
That is common difference a₂ – a₁ = a₃ – a₂ = a₄ – a₃ … is the same every time.
So, the given list of numbers forms an A.P with the common difference d = a.
∴ Next three terms are 5a, 6a, 7a,…

xi) Given a, a², a³, a⁴,……..
We have common difference,
a₂ – a₁ = a² – a = a(a – 1)
a₃ – a₂ = a³ – a² = a² (a – 1)
a₄ – a₃ = a⁴ – a³ = a³(a – 1)
That is common difference a₂ – a₁ ≠ a₃ – a₂ ≠ a₄ ≠ – a₃ … is not same every time.
So, the given list of numbers does not form an A. P.

xii) Given \(\sqrt{2}\), \(\sqrt{8}\), \(\sqrt{18}\), \(\sqrt{32}\),…
We have common difference,

That is common difference a₂ – a₁ = a₃ – a₂ = a₄ – a₃ … is the same every time.
So, the given list of numbers forms an A.P with the common difference d = \(\sqrt{2}\).
∴ Next three terms are
\(\sqrt{50}\), \(\sqrt{72}\), \(\sqrt{98}\),…

xiii) Given \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{9}\), \(\sqrt{12}\),…
We have common difference,
a₂ – a₁ = \(\sqrt{6}\) – \(\sqrt{3}\)
a₃ – a₂ = \(\sqrt{9}\) – \(\sqrt{6}\)
a₄ – a₃ = \(\sqrt{12}\) – \(\sqrt{9}\)
That is common difference a₂ – a₁ ≠ a₃ – a₂ ≠ a₄ – a₃ … is not same every time.
So, the given list of numbers does not form an A.P.

xiv) Given 1², 3², 5², 7²,…
We have common difference,
a₂ – a₁ = 3² – 1² = 9 – 1 = 8
a₃ – a₂ = 5² – 3² = 25 – 9 = 16
a₄ – a₃ = 7² – 5² = 49 – 25 = 24
That is common difference a₂ – a₁ ≠ a₃ – a₂ ≠ a₄ – a₃ … is not same every time.
So, the given list of numbers does not form an A.P.

xv) Given 1², 5², 7², 73,…
We have common difference,
a₂ – a₁ = 5² – 1² = 25 – 1 = 24
a₃ – a₂ = 7² – 5² = 49 – 25 = 24
a₄ – a₃ = 73 – 7² = 73 – 49 = 24
That is common difference a₂ – a₁ = a₃ – a₂ = a₄ – a₃ … is the same every time.
So, the given list of numbers forms an A.P with the common difference d = 24.
Next three terms are 73 + 24 = 97
97 + 24 = 121
121 + 24 = 145
Therefore, 97, 121, 145,……..