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**Ques 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 â‚¹ 15 for the first km and â‚¹ 8 for each additional km.****(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.****(iii) The cost of digging a well after every metre of digging, when it costs â‚¹ 150 for the first metre and rises by â‚¹ 50 for each subsequent metre.****(iv) The amount of money in the account every year, when â‚¹ 10,000 is deposited at compound interest at 8% per annum.****Sol:** (i) Let us consider,

The first term (T_{1}) = Fare for the first km = â‚¹ 15 since, the taxi fare beyond the first km is â‚¹ 8 for each additional km. â‡’ T_{1} = 15

âˆ´ Fare for 2 km = â‚¹15 + 1 Ã— â‚¹ 8 â‡’ T_{2} = a + 8 [where a = 15]

Fare for 3 km = â‚¹ 15 + 2 Ã— â‚¹ 8 â‡’ T_{3} = a + 16

Fare for 4 km = â‚¹ 15 + 3 Ã— â‚¹ 8 â‡’ T_{4} = a + 24

Fare for 5 km = â‚¹ 15 + 4 Ã— â‚¹ 8 â‡’ T_{5} = a + 32

Fare for n km = â‚¹ 15 + (n - 1) 8 â‡’ Tn = a + (n - 1) 8

We see that above terms form an A.P.

(ii) Let the amount of air in the cylinder = x

âˆ´ Air removed in 1st stroke =

â‡’ Air left after 1st stroke

Air left after 2nd stroke

Air left after 3rd stroke

Air left after 4th stroke

Thus, the terms are:

Here,

Since The above terms are not in A.P.

(iii) Here, The cost of digging for 1st metre = â‚¹ 150

The cost of digging for first 2 metres = â‚¹ 150 + â‚¹ 50 = â‚¹ 200

The cost of digging for first 3 metres = â‚¹ 150 + (â‚¹ 50) Ã— 2 = â‚¹ 250

The cost of digging for first 4 metres = â‚¹ 150 + (â‚¹ 50) Ã— 3 = â‚¹ 300

âˆ´ The terms are: 150, 200, 250, 300, ...

Since, 200 - 150 = 50

And 250 - 200 = 50

â‡’ (200 - 150) = (250 - 200)

âˆ´The above terms form an A.P.

(iv) âˆµ The amount at the end of 1st year

The amount at the end of 2nd year =

The amount at the end of 3rd year =

The amount at the end of 4th year =

âˆ´ The terms are

Obviously,

âˆ´ The above terms are not in A.P.**Ques ****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 = 0****(v) a = - 1.25, d = - 0.25****Sol:** (i) âˆµ T_{n} = a + (n - 1) d

âˆ´ For a = 10 and d = 10, we have:

T_{1} = 10 + (1 - 1) Ã— 10 = 10 + 0 = 10

T_{2} = 10 + (2 - 1) Ã— 10 = 10 + 10 = 20

T_{3} = 10 + (3 - 1) Ã— 10 = 10 + 20 = 30

T_{4} = 10 + (4 - 1) Ã— 10 = 10 + 30 = 40

Thus, the first four terms of A.P. are:

10, 20, 30, 40.

(ii) âˆµ T_{n} = a + (n - 1) d

âˆ´ For a = - 2 and d = 0, we have:

T_{1} = -2 + (1 - 1) Ã— 0 = -2 + 0 = -2

T_{2} = -2 + (2 - 1) Ã— 0 = -2 + 0 = -2

T_{3} = -2 + (3 - 1) Ã— 0 = -2 + 0 = -2

T_{4} = -2 + (4 - 1) Ã— 0 = -2 + 0 = -2

âˆ´ The first four terms are:

- 2, - 2, - 2, - 2.

(iii) âˆµ T_{n} = a + (n - 1) d

âˆ´ For a = 4 and d = - 3, we have:

T_{1} = 4 + (1 - 1) Ã— (- 3) = 4 + 0 = 4

T_{2} = 4 + (2 - 1) Ã— (- 3) = 4 + (- 3) = 1

T_{3} = 4 + (3 - 1) Ã— (- 3) = 4 + (- 6) = - 2

T_{4} = 4 + (4 - 1) Ã— (- 3) = 4 + (- 9) = - 5

Thus, the first four terms are:

4, 1, - 2, - 5.

(iv) âˆµ T_{n} = a + (n - 1) d

For a = - 1 and d = 1/2, we get

âˆ´ The first four terms are:

(v) âˆµ T_{n} = a + (n - 1) d

For a = - 1.25 and d = - 0.25, we get

T_{1} = - 1.25 + (1 - 1) Ã— (- 0.25) = - 1.25 + 0 = - 1.25

T_{2} = - 1.25 + (2 - 1) Ã— (- 0.25) = - 1.25 + (- 0.25) = - 1.50

T_{3} = - 1.25 + (3 - 1) Ã— (- 0.25) = - 1.25 + (- 0.50) = - 1.75

T_{4 }= - 1.25 + (4 - 1) Ã— (- 0.25) = - 1.25 + (- 0.75) = - 2.0

Thus, the four terms are:

- 1.25, - 1.50, - 1.75, - 2.0**Ques ****3: For the following APs, write the first term and the common difference:****(i) 3, 1, - 1, - 3, ... ****(ii) - 5, - 1, 3, 7, ...****(iii) ... ****(iv) 0.6, 1.7, 2.8, 3.9, ... ****Sol:** (i) We have: 3, 1, - 1, - 3, .....

â‡’ T_{1} = 3 â‡’ a = 3

T_{2} = 1

T_{3} = - 1

T_{4} = - 3

âˆ´ T_{2} - T_{1} = 1 - 3 = - 2

T_{4} - T_{3} = - 3 - (- 1) = - 3 + 2 = - 2 â‡’ d = - 2

Thus, a = 3 and d = - 2

(ii) We have: - 5, - 1, 3, 7, .....

â‡’

Thus, a = -5 and d = 4

(iii) We have: .....

â‡’

Thus, a =1/3 and d = 4/3

(iv) We have: 0.6, 1.7, 2.8, 3.9, .....

â‡’ T_{1} = 0.6â‡’ a = 0.6

T_{2} = 1.7 â‡’ d = T_{2} - T_{1} = 1.7 - 0.6 = 1.1

T_{3} = 2.8

T_{4} = 3.9 â‡’ d = T_{4} - T_{3} = 3.9 - 2.8 = 1.1

Thus, a = 0.6 and d = 1.1**Ques ****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) **** ...****(iii) - 1.2, - 3.2, - 5.2, - 7.2, ... ****(iv) - 10, - 6, - 2, 2, ...****(v) ****(vi) 0.2, 0.22, 0.222, 0.2222, ... ****(vii) 0, - 4, - 8, - 12, ... ****(viii)****(ix) 1, 3, 9, 27, ... ****(x) a, 2a, 3a, 4a, ...(xi) a, a ^{2}, a^{3}, a^{4}, ... **

(xiii) âˆš3, âˆš6 âˆš9, âˆš12, ...

âˆ´ T

âˆ´ The given numbers do not form an A.P.

(ii) We have: , .....

âˆ´T

T

T

T

âˆµ T

âˆ´The given numbers form an A.P.

âˆ´T

T

T

Thus, d = 1/2 and T

(iii) We have: - 1.2, - 3.2, - 5.2, - 7.2, .....

âˆ´T

T2 - T1 = -3.2 + 1.2 = -2

T3 - T2 = -5.2 + 3.2 = -2

T4 - T3 = -7.2 + 5.2 = -2

âˆµ T2 - T

âˆ´ The given numbers form an A.P.

Such that d = - 2.

Now, T

T

T

Thus, d = - 2 and T

(iv) We have: - 10, - 6, - 2, 2, .....

âˆ´ T

T

T

T

âˆµ T

âˆ´The given numbers form an A.P.

Now, T

T

T

Thus, d = 4 and T

(v) We have:

âˆ´

â‡’ The given numbers form an A.P.

Now, T

= 3+3âˆš2+âˆš2 = 3+4âˆš2

T

= 3+4âˆš2+âˆš2 = 3+5âˆš2

T

= 3+5âˆš2+âˆš2 = 3+6âˆš2

Thus, d = âˆš2 and T

(vi) We have: 0.2, 0.22, 0.222, 0.2222, .....

Since,

T

âˆ´ The given numbers do not form an A.P.

(vii) We have: 0, - 4, - 8, - 12, .....

âˆ´ T

T

T

T

âˆµ T

âˆ´The given numbers form an A.P.

Now, T

T

T

Thus, d = - 4 and T

(viii) We have:

âˆ´

T

T

T

âˆµ T

âˆ´ The given number form an A.P.

Now,

T

T

T

Thus,

(ix) We have: 1, 3, 9, 27, .....

Here,

âˆµ T

âˆ´The given numbers do not form an A.P.

(x) We have: a, 2a, 3a, 4a, .....

âˆ´ T

T

T

T

âˆµ T

âˆ´The numbers form an A.P.

Now, T

T

T

Thus, d = a and T

(xi) We have: a, a

Since,

T_{2} - T_{1} â‰ T_{4} - T_{3}

âˆ´ The given terms are not in A.P.

(xii) We have:

âˆ´

âˆ´ The given numbers form an A.P.

Now,

Thus,

(xiii) We have:

âˆ´ T_{2} - T_{1} â‰ T_{4} - T_{3}

â‡’ The given terms do not form an A.P.

(xiv) We have: 1^{2}, 3^{2}, 5^{2}, 7^{2}, .....

âˆµ T_{2} âˆ’ T_{1} â‰ T_{4} âˆ’ T_{3}

âˆ´ The given terms do not form an A.P.

(xv) We have: 1^{2}, 5^{2}, 7^{2}, 7^{3}, .....

âˆ´ T_{1} = 1^{2} = 1, T_{2} = 5^{2} = 25, T_{3} = 7^{2} = 49, T_{4} = 73

T_{2} - T_{1} = 25 - 1 = 24

T_{3} - T_{2} = 49 - 25 = 24

T_{4} - T_{3} = 73 - 49 = 24

âˆµ T_{2} - T_{1} = T_{3} - T_{2} = T_{4} - T_{3} = 24 â‡’ d = 24

âˆ´ The numbers form an A.P.

Now, T_{5} = T_{4} + 24 = 73 + 24 = 97

T_{6} = T_{5} + 24 = 97 + 24 = 121

T_{7} = T_{6} + 24 = 121 + 24 = 145

Thus, d = 24 and T_{5 }= 97, T_{6} = 121, T_{7} = 145**nth Term of an A.P.**

The nth term T_{n} of the A.P. with first term â€˜aâ€™ and common difference â€˜dâ€™ is given by

T_{n} = a + (n - 1) d

â€˜T_{n}â€™ is also called the general term of the A.P. If there are â€˜mâ€™ terms in the A.P., then â€˜T_{m}â€™ represents the last term which is generally denoted by â€˜lâ€™.

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