(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.

Solution:

Let the taxi fare for the first km be a₁ = 15.

Fare for 2 km = 15 + 8 = 23

Fare for 3 km = 23 + 8 = 31

Fare for 4 km = 31 + 8 = 39

The list of numbers obtained is 15, 23, 31, 39, ...

Here, the difference between consecutive terms is constant:

23 − 15 = 8

31 − 23 = 8

39 − 31 = 8

Since the difference between successive terms is always the same (8), this situation forms an Arithmetic Progression.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Solution:

Let the initial volume of air present in the cylinder be V.

Air removed the first time = (1/4)V

Remaining air, a₂ = V − (1/4)V = (3/4)V

Air removed the second time = (1/4)(3/4)V = (3/16)V

Remaining air, a₃ = (3/4)V − (3/16)V = (9/16)V = (3/4)²V

The list of remaining volumes is V, (3/4)V, (3/4)²V, ...

Let us check the difference between consecutive terms:

a₂ − a₁ = (3/4)V − V = −(1/4)V

a₃ − a₂ = (9/16)V − (3/4)V = −(3/16)V

Since a₂ − a₁ ≠ a₃ − a₂, the common difference is not the same. Therefore, this situation does not form an Arithmetic Progression.

(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.

Solution:

Cost of digging for the first metre, a₁ = 150

Cost of digging for 2 metres, a₂ = 150 + 50 = 200

Cost of digging for 3 metres, a₃ = 200 + 50 = 250

Cost of digging for 4 metres, a₄ = 250 + 50 = 300

The list of numbers obtained is 150, 200, 250, 300, ...

Checking the consecutive differences:

200 − 150 = 50

250 − 200 = 50

300 − 250 = 50

Since the difference between any two successive terms is constant (50), this situation forms an Arithmetic Progression.

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

Solution:

We know that the amount under compound interest is given by the formula A = P(1 + r/100)ⁿ.

Amount in the account at the start, a₁ = 10000

Amount after 1 year, a₂ = 10000(1 + 8/100)¹ = 10000(1.08)

Amount after 2 years, a₃ = 10000(1 + 8/100)² = 10000(1.08)²

Amount after 3 years, a₄ = 10000(1 + 8/100)³ = 10000(1.08)³

The list of amounts across subsequent years is 10000, 10000(1.08), 10000(1.08)², 10000(1.08)³, ...

Let us check the differences:

a₂ − a₁ = 10000(1.08 − 1) = 800

a₃ − a₂ = 10000(1.08)(1.08 − 1) = 864

Since the difference between successive years increases because the interest is calculated on a larger base amount each year, the common difference is not uniform. Therefore, this situation does not form an Arithmetic Progression.

(i) a = 10, d = 10

Solution:

First term, a₁ = a = 10

Second term, a₂ = a + d = 10 + 10 = 20

Third term, a₃ = a₂ + d = 20 + 10 = 30

Fourth term, a₄ = a₃ + d = 30 + 10 = 40

Hence, the first four terms of the AP are 10, 20, 30, and 40.

(ii) a = −2, d = 0

Solution:

First term, a₁ = a = −2

Second term, a₂ = a + d = −2 + 0 = −2

Third term, a₃ = a₂ + d = −2 + 0 = −2

Fourth term, a₄ = a₃ + d = −2 + 0 = −2

Hence, the first four terms of the AP are −2, −2, −2, and −2.

(iii) a = 4, d = −3

Solution:

First term, a₁ = a = 4

Second term, a₂ = a + d = 4 + (−3) = 1

Third term, a₃ = a₂ + d = 1 + (−3) = −2

Fourth term, a₄ = a₃ + d = −2 + (−3) = −5

Hence, the first four terms of the AP are 4, 1, −2, and −5.

(iv) a = −1, d = 1/2

Solution:

First term, a₁ = a = −1

Second term, a₂ = a + d = −1 + 1/2 = −1/2

Third term, a₃ = a₂ + d = −1/2 + 1/2 = 0

Fourth term, a₄ = a₃ + d = 0 + 1/2 = 1/2

Hence, the first four terms of the AP are −1, −1/2, 0, and 1/2.

(v) a = −1.25, d = −0.25

Solution:

First term, a₁ = a = −1.25

Second term, a₂ = a + d = −1.25 + (−0.25) = −1.50

Third term, a₃ = a₂ + d = −1.50 + (−0.25) = −1.75

Fourth term, a₄ = a₃ + d = −1.75 + (−0.25) = −2.00

Hence, the first four terms of the AP are −1.25, −1.50, −1.75, and −2.00.

(i) 3, 1, −1, −3, ...

Solution:

The given arithmetic progression is 3, 1, −1, −3, ...

First term (a) = 3

Common difference (d) = Second term − First term = 1 − 3 = −2

Hence, first term a = 3 and common difference d = −2.

(ii) −5, −1, 3, 7, ...

Solution:

The given arithmetic progression is −5, −1, 3, 7, ...

First term (a) = −5

Common difference (d) = Second term − First term = −1 − (−5) = −1 + 5 = 4

Hence, first term a = −5 and common difference d = 4.

(iii) 1/3, 5/3, 9/3, 13/3, ...

Solution:

The given arithmetic progression is 1/3, 5/3, 9/3, 13/3, ...

First term (a) = 1/3

Common difference (d) = Second term − First term = 5/3 − 1/3 = 4/3

Hence, first term a = 1/3 and common difference d = 4/3.

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

Solution:

The given arithmetic progression is 0.6, 1.7, 2.8, 3.9, ...

First term (a) = 0.6

Common difference (d) = Second term − First term = 1.7 − 0.6 = 1.1

Hence, first term a = 0.6 and common difference d = 1.1.

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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, ...

Solution:

Let the terms be a₁ = 2, a₂ = 4, a₃ = 8, and a₄ = 16.

Finding the differences between consecutive terms:

a₂ − a₁ = 4 − 2 = 2

a₃ − a₂ = 8 − 4 = 4

Since a₂ − a₁ ≠ a₃ − a₂, the difference between consecutive terms is not constant. Therefore, the given list of numbers does not form an AP.

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(ii) 2, 5/2, 3, 7/2, ...

Solution:

Let the terms be a₁ = 2, a₂ = 5/2, a₃ = 3, and a₄ = 7/2.

Finding the differences between consecutive terms:

a₂ − a₁ = 5/2 − 2 = 1/2

a₃ − a₂ = 3 − 5/2 = 1/2

a₄ − a₃ = 7/2 − 3 = 1/2

Since the difference between consecutive terms is constant, the given list of numbers forms an AP with a common difference d = 1/2.

The next three terms are:

a₅ = 7/2 + 1/2 = 4

a₆ = 4 + 1/2 = 9/2

a₇ = 9/2 + 1/2 = 5

Hence, the common difference is 1/2 and the next three terms are 4, 9/2, and 5.

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(iii) −1.2, −3.2, −5.2, −7.2, ...

Solution:

Let the terms be a₁ = −1.2, a₂ = −3.2, a₃ = −5.2, and a₄ = −7.2.

Finding the differences between consecutive terms:

a₂ − a₁ = −3.2 − (−1.2) = −2.0

a₃ − a₂ = −5.2 − (−3.2) = −2.0

a₄ − a₃ = −7.2 − (−5.2) = −2.0

Since the difference between consecutive terms is constant, the given list of numbers forms an AP with a common difference d = −2.

The next three terms are:

a₅ = −7.2 + (−2) = −9.2

a₆ = −9.2 + (−2) = −11.2

a₇ = −11.2 + (−2) = −13.2

Hence, the common difference is −2 and the next three terms are −9.2, −11.2, and −13.2.

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(iv) −10, −6, −2, 2, ...

Solution:

Let the terms be a₁ = −10, a₂ = −6, a₃ = −2, and a₄ = 2.

Finding the differences between consecutive terms:

a₂ − a₁ = −6 − (−10) = 4

a₃ − a₂ = −2 − (−6) = 4

a₄ − a₃ = 2 − (−2) = 4

Since the difference between consecutive terms is constant, the given list of numbers forms an AP with a common difference d = 4.

The next three terms are:

a₅ = 2 + 4 = 6

a₆ = 6 + 4 = 10

a₇ = 10 + 4 = 14

Hence, the common difference is 4 and the next three terms are 6, 10, and 14.

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(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, ...

Solution:

Let the terms be a₁ = 3, a₂ = 3 + √2, a₃ = 3 + 2√2, and a₄ = 3 + 3√2.

Finding the differences between consecutive terms:

a₂ − a₁ = (3 + √2) − 3 = √2

a₃ − a₂ = (3 + 2√2) − (3 + √2) = √2

a₄ − a₃ = (3 + 3√2) − (3 + 2√2) = √2

Since the difference between consecutive terms is constant, the given list of numbers forms an AP with a common difference d = √2.

The next three terms are:

a₅ = (3 + 3√2) + √2 = 3 + 4√2

a₆ = (3 + 4√2) + √2 = 3 + 5√2

a₇ = (3 + 5√2) + √2 = 3 + 6√2

Hence, the common difference is √2 and the next three terms are 3 + 4√2, 3 + 5√2, and 3 + 6√2.

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

Solution:

Let the terms be a₁ = 0.2, a₂ = 0.22, a₃ = 0.222, and a₄ = 0.2222.

Finding the differences between consecutive terms:

a₂ − a₁ = 0.22 − 0.2 = 0.02

a₃ − a₂ = 0.222 − 0.22 = 0.002

Since a₂ − a₁ ≠ a₃ − a₂, the difference between consecutive terms is not constant. Therefore, the given list of numbers does not form an AP.

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(vii) 0, −4, −8, −12, ...

Solution:

Let the terms be a₁ = 0, a₂ = −4, a₃ = −8, and a₄ = −12.

Finding the differences between consecutive terms:

a₂ − a₁ = −4 − 0 = −4

a₃ − a₂ = −8 − (−4) = −4

a₄ − a₃ = −12 − (−8) = −4

Since the difference between consecutive terms is constant, the given list of numbers forms an AP with a common difference d = −4.

The next three terms are:

a₅ = −12 + (−4) = −16

a₆ = −16 + (−4) = −20

a₇ = −20 + (−4) = −24

Hence, the common difference is −4 and the next three terms are −16, −20, and −24.

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(viii) −1/2, −1/2, −1/2, −1/2, ...

Solution:

Let the terms be a₁ = −1/2, a₂ = −1/2, a₃ = −1/2, and a₄ = −1/2.

Finding the differences between consecutive terms:

a₂ − a₁ = −1/2 − (−1/2) = 0

a₃ − a₂ = −1/2 − (−1/2) = 0

a₄ − a₃ = −1/2 − (−1/2) = 0

Since the difference between consecutive terms is constant, the given list of numbers forms an AP with a common difference d = 0.

The next three terms remain unchanged because the common difference is zero:

a₅ = −1/2

a₆ = −1/2

a₇ = −1/2

Hence, the common difference is 0 and the next three terms are −1/2, −1/2, and −1/2.

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

Solution:

Let the terms be a₁ = 1, a₂ = 3, a₃ = 9, and a₄ = 27.

Finding the differences between consecutive terms:

a₂ − a₁ = 3 − 1 = 2

a₃ − a₂ = 9 − 3 = 6

Since a₂ − a₁ ≠ a₃ − a₂, the difference between consecutive terms is not constant. Therefore, the given list of numbers does not form an AP.

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

Solution:

Let the terms be a₁ = a, a₂ = 2a, a₃ = 3a, and a₄ = 4a.

Finding the differences between consecutive terms:

a₂ − a₁ = 2a − a = a

a₃ − a₂ = 3a − 2a = a

a₄ − a₃ = 4a − 3a = a

Since the difference between consecutive terms is constant, the given list of numbers forms an AP with a common difference d = a.

The next three terms are:

a₅ = 4a + a = 5a

a₆ = 5a + a = 6a

a₇ = 6a + a = 7a

Hence, the common difference is a and the next three terms are 5a, 6a, and 7a.

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(xi) a, a², a³, a⁴, ...

Solution:

Let the terms be a₁ = a, a₂ = a², a₃ = a³, and a₄ = a⁴.

Finding the differences between consecutive terms:

a₂ − a₁ = a² − a = a(a − 1)

a₃ − a₂ = a³ − a² = a²(a − 1)

Since a₂ − a₁ ≠ a₃ − a₂, the difference between consecutive terms is not constant. Therefore, the given list of numbers does not form an AP.

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(xii) √2, √8, √18, √32, ...

Solution:

The given terms can be rewritten as:

a₁ = √2

a₂ = √8 = √(4 × 2) = 2√2

a₃ = √18 = √(9 × 2) = 3√2

a₄ = √32 = √(16 × 2) = 4√2

Finding the differences between consecutive terms:

a₂ − a₁ = 2√2 − √2 = √2

a₃ − a₂ = 3√2 − 2√2 = √2

a₄ − a₃ = 4√2 − 3√2 = √2

Since the difference between consecutive terms is constant, the given list of numbers forms an AP with a common difference d = √2.

The next three terms are:

a₅ = 4√2 + √2 = 5√2 = √(25 × 2) = √50

a₆ = 5√2 + √2 = 6√2 = √(36 × 2) = √72

a₇ = 6√2 + √2 = 7√2 = √(49 × 2) = √98

Hence, the common difference is √2 and the next three terms are √50, √72, and √98.

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(xiii) √3, √6, √9, √12, ...

Solution:

The given terms can be simplified as a₁ = √3, a₂ = √6, a₃ = 3, and a₄ = 2√3.

Finding the differences between consecutive terms:

a₂ − a₁ = √6 − √3 = √3(√2 − 1)

a₃ − a₂ = 3 − √6 = √3(√3 − √2)

Since a₂ − a₁ ≠ a₃ − a₂, the difference between consecutive terms is not constant. Therefore, the given list of numbers does not form an AP.

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(xiv) 1², 3², 5², 7², ...

Solution:

The squares can be simplified to absolute values:

a₁ = 1, a₂ = 9, a₃ = 25, and a₄ = 49.

Finding the differences between consecutive terms:

a₂ − a₁ = 9 − 1 = 8

a₃ − a₂ = 25 − 9 = 16

Since a₂ − a₁ ≠ a₃ − a₂, the difference between consecutive terms is not constant. Therefore, the given list of numbers does not form an AP.

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(xv) 1², 5², 7², 73, ...

Solution:

The terms can be simplified to numerical values:

a₁ = 1, a₂ = 25, a₃ = 49, and a₄ = 73.

Finding the differences between consecutive terms:

a₂ − a₁ = 25 − 1 = 24

a₃ − a₂ = 49 − 25 = 24

a₄ − a₃ = 73 − 49 = 24

Since the difference between consecutive terms is constant, the given list of numbers forms an AP with a common difference d = 24.

The next three terms are:

a₅ = 73 + 24 = 97

a₆ = 97 + 24 = 121

a₇ = 121 + 24 = 145

Hence, the common difference is 24 and the next three terms are 97, 121, and 145.