160-NCERT-10-5-Arithmetic Progressions - Ex-5.1

NCERT

10th Mathematics

Exercise 5.1

Topic: 5 Arithmetic Progressions

Click here for ⇨ NCERT-10-4-Quadratic Equations-Ex- 4.4

EXERCISE 5.1

Q1. 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 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 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.

Explanation:

1) An arithmetic progression is a list of numbers in which each term is obtained by

adding a fixed number to the preceding term except the first term.

2) The fixed number is known as the common difference of an AP.

3) The common difference may be positive, negative, or zero.

4) If the first of an AP is "a" and the common difference is "d" then the terms of an

AP will be:  a, (a + d), (a + 2d), (a + 3d) ...

Solution:

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

1) According to the problem,

a) Taxi fare for the first km = 15.

b) Taxi fare for the first 2 km = 15 + 8 = 23.

c) Taxi fare for the first 3 km = 15 + 8 + 8 = 31.

2) As the terms increment by a constant 8, it forms an AP.

(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.

1) Let the amount of air in the cylinder be x.

2) According to the problem, every time the vacuum pump removes (1/4) of air

remaining in the cylinder, so

a) The volume after first removal

= x - (x/4)

= x(1 - 1/4)

= 3x/4 ---------------- equation 1

b) The volume after second removal 

= 3x/4 - 1/4(3x/4)

= (3x/4)(1 - 1/4)

= (3x/4)(4 - 1)/4

= (3x/4)(3/4)

= 9x/16 ---------------- equation 2

c) The volume after third removal 

= 9x/16 - 1/4(9x/16)

= (9x/16)(1 - 1/4)

= (9x/16)(3/4)

= 27x/64 ---------------- equation 3

3) Now we will check the terms x, 3x/4, 9x/16, 27x/64 are in AP or not.

a) Second term - first term = 3x/4 - x 

   = (3x - 4x)/4

   = - x/4 ---------------- equation 4.

b) Third term - Second term = 9x/16 - 3x/4 

     = (9x - 12x)/16

     = - 3x/16 ---------------- equation 5.

4) From equation 4 and equation 5, as the differences between the terms are not

same, we can say that these terms are not in AP.

(iii) The cost of digging a well after every meter of digging, when it costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter. 

1) According to the problem,

a) Cost of digging the well after 1 meter = 150.

b) Cost of digging the well after 2 meter = 150 + 50 = 200.

c) Cost of digging the well after 3 meter = 150 + 50 + 50 = 250.

2) As the terms increment by a constant 50, it forms an AP.

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

1) We know that if Rs P is invested at the rate of r % per annum for n years, the

amount received will be A = P[1+(r/100)]ⁿ.

2) According to the problem, for P = 10000, r = 8 %, 

a) The amount after first year

A = P[1+(r/100)]ⁿ

A = 10000[1+(8/100)]¹

A = 10000[1+(8/100)] ---------------- equation 1

b) The amount after second year 

A = P[1+(r/100)]ⁿ

A = 10000[1+(8/100)]² ---------------- equation 2

c) The amount after third year 

A = P[1+(r/100)]ⁿ

A = 10000[1+(8/100)]³ ---------------- equation 3

3) Now we will check the terms [1+(8/100)], [1+(8/100)]², [1+(8/100)]³, 

[1+(8/100)]⁴ are in AP or not.

a) Second term - first term

= 10000[1+(8/100)]² - 10000[1+(8/100)]

= 10000[1+(8/100)]{1+(8/100) - 1}

= 10000[1+(8/100)](8/100) 

= 10000(8/100)[1+(8/100)] ---------------- equation 4.

b) Third term - Second term 

= 10000[1+(8/100)]³ - 10000[1+(8/100)]²

= 10000[1+(8/100)]²{1+(8/100) - 1}

= 10000[1+(8/100)]²(8/100) 
= 10000(8/100)[1+(8/100)]² ---------------- equation 5.

4) From equation 4 and equation 5, as the differences between the terms are not

same, we can say that these terms are not in AP.

Q2. Write the 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 = 1/2     (v) a = – 1.25, d = – 0.25

Explanation:

1) An arithmetic progression is a list of numbers in which each term is obtained by

adding a fixed number to the preceding term except the first term.

2) The fixed number is known as the common difference of an AP.

3) The common difference of may positive, negative, or zero.

4) If the first of an AP is "a" and the common difference is "d" then the terms of an

AP will be:  a, (a + d), (a + 2d), (a + 3d) ...

Solution:

(i) a = 10, d = 10

1) Let the terms of an AP be a₁, a₂, a₃, a_{4.
2) Here,} a₁ = a = 10 is the first term and d = 10 is the common difference.
3) Now we will find the terms of an AP:

a) First term:

a₁ = 10

b) Second term:

a₂ = a₁ + d

a₂ = 10 + 10
a₂ = 20

c) Third term:

a₃ = a₂ + d
a₃ = 20 + 10
a₃ = 30

d) Fourth term:

a₄ = a₃ + d
a₄ = 30 + 10
a₄ = 40

4) So the first 4 terms of an AP with a =10 and d = 10 are 10, 20, 30, and 40.

(ii) a = –2, d = 0

1) Let the terms of an AP be a₁, a₂, a₃, a_{4.
2) Here,} a₁ = a = - 2 is the first term and d = 0 is the common difference.
3) Now we will find the terms of an AP:

a) First term:

a₁ = - 2

b) Second term:

a₂ = a₁ + d

a₂ = - 2 + 0
a₂ = - 2

c) Third term:

a₃ = a₂ + d
a₃ = - 2 + 0
a₃ = - 2

d) Fourth term:

a₄ = a₃ + d
a₄ = - 2 + 0
a₄ = - 2

4) So first 4 terms of an AP with a = - 2 and d = 0 are - 2, - 2, - 2, and - 2.

(iii) a = 4, d = – 3

1) Let the terms of an AP be a₁, a₂, a₃, a_{4.
2) Here,} a₁ = a = 4 is the first term and d = - 3 is the common difference.
3) Now we will find the terms of an AP:

a) First term:

a₁ = 4

b) Second term:

a₂ = a₁ + d

a₂ = 4 - 3
a₂ = 1

c) Third term:

a₃ = a₂ + d
a₃ = 1 - 3
a₃ = - 2

d) Fourth term:

a₄ = a₃ + d
a₄ = - 2 - 3
a₄ = - 5

4) So first 4 terms of an AP with a = 4 and d = - 3 are 4, 1, - 2, and - 5.

(iv) a = – 1, d = 1/2

1) Let the terms of an AP be a₁, a₂, a₃, a_{4.
2) Here,} a₁ = a = - 1 is the first term and d = 1/2 is the common difference.
3) Now we will find the terms of an AP:

a) First term:

a₁ = - 1

b) Second term:

a₂ = a₁ + d

a₂ = - 1 + 1/2
a₂ = - 1/2

c) Third term:

a₃ = a₂ + d
a₃ = - 1/2 + 1/2
a₃ = 0

d) Fourth term:

a₄ = a₃ + d
a₄ = 0 + 1/2
a₄ = 1/2

4) So first 4 terms of an AP with a = - 1 and d = 1/2 are - 1, - 1/2, 0, and 1/2.

(v) a = – 1.25, d = – 0.25

1) Let the terms of an AP be a₁, a₂, a₃, a_{4.
2) Here,} a₁ = a = - 1.25 is the first term and d = - 0.25 is the common difference.
3) Now we will find the terms of an AP:

a) First term:

a₁ = - 1.25

b) Second term:

a₂ = a₁ + d

a₂ = - 1.25 - 0.25
a₂ = - 1.50

c) Third term:

a₃ = a₂ + d
a₃ = - 1.50 - 0.25
a₃ = - 1.75

d) Fourth term:

a₄ = a₃ + d
a₄ = - 1.75 - 0.25
a₄ = - 2.00

4) So first 4 terms of an AP with a = - 1.25 and d = - 0.25 are - 1.25, - 1.50, - 1.75,

and - 2.00.

Q3. For the following APs, write the first term and the common difference:

(i) 3, 1, – 1, – 3, . . .     (ii) – 5, – 1, 3, 7, . . .

 

(iii) 1/3, 5/3, 9/3, 13/3, . . .     (iv) 0.6, 1.7, 2.8, 3.9, . . .

Explanation:

1) An arithmetic progression is a list of numbers in which each term is obtained by

adding a fixed number to the preceding term except the first term.

2) The fixed number is known as the common difference of an AP.

3) The common difference may be positive, negative, or zero.

4) If the first of an AP is "a" and the common difference is "d" then the terms of an

AP will be:  a, (a + d), (a + 2d), (a + 3d) ...

5) For the terms a₁, a₂, a₃, a_4, when a₂ - a₁ = a₃ - a₂ = a₄ - a₃ = d, then we say

that the terms a₁, a₂, a₃, a_4, are in AP.

Solution:

(i) 3, 1, – 1, – 3, . . .

1) Here the first term is a = 3.

2) Here a₁ = a = 3, a₂ = 1, a₃ = - 1, a₄ = - 3.

3) Here, 

a) First difference:

d = a₂ - a₁ = 1 - 3

d = a₂ - a₁ = - 2 --------- equation 1

b) Second difference:

d = a₃ - a₂ = - 1 - 1

d = a₃ - a₂ = - 2 --------- equation 2

4) Here the first term a = 3 and the common difference d = - 2.  

(ii) – 5, – 1, 3, 7, . . .

1) Here the first term is a = - 5.

2) Here a₁ = a = - 5, a₂ = - 1, a₃ = 3, a₄ = 7.

3) Here, 

a) First difference:

d = a₂ - a₁ = - 1 - (- 5)

d = a₂ - a₁ = - 1 + 5 

d = a₂ - a₁ = 4 --------- equation 1

b) Second difference:

d = a₃ - a₂ = 3 - (- 1)

d = a₃ - a₂ = 3 + 1 

d = a₃ - a₂ = 4 --------- equation 2

4) Here the first term a = - 5 and the common difference d = 4.

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

1) Here the first term is a = 1/3.

2) Here a₁ = a = 1/3, a₂ = 5/3, a₃ = 9/3, a₄ = 13/3.

3) Here, 

a) First difference:

d = a₂ - a₁ = 5/3 - 1/3

d = a₂ - a₁ = (5 - 1)/3 

d = a₂ - a₁ = 4/3 --------- equation 1

b) Second difference:

d = a₃ - a₂ = 9/3 - 5/3

d = a₃ - a₂ = (9 - 5)/3 

d = a₃ - a₂ = 4/3 --------- equation 2

4) Here the first term a = 1/3 and the common difference d = 4/3.

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

1) Here the first term is a = 0.6.

2) Here a₁ = a = 0.6, a₂ = 1.7, a₃ = 2.8, a₄ = 3.9.

3) Here, 

a) First difference:

d = a₂ - a₁ = 1.7 - 0.6

d = a₂ - a₁ = 1.1 --------- equation 1

b) Second difference:

d = a₃ - a₂ = 2.8 - 1.7

d = a₃ - a₂ = 1.1 --------- equation 2

4) Here the first term a = 0.6 and the common difference d = 1.1.

Q4. 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, 5/2, 3, 7/2, . . .      

(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . 

(iv) – 10, – 6, – 2, 2, . . .

(v) 3, 3 + √2 , 3 + 2 √2 , 3 + 3 √2 , . . .

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

(vii) 0, – 4, – 8, –12, . . .

(viii) - 1/2, - 1/2, - 1/2, - 1/2, . . .

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

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

(xi) a, a², a³, a⁴, . . .

(xii) √2, √8, √18 , √32, . . .

(xiii) √3, √6, √9 , √12 , . . .

(xiv) 1², 3², 5², 7², . . .

(xv) 1², 5², 7², 7³, . . .

Explanation:

1) An arithmetic progression is a list of numbers in which each term is obtained by

adding a fixed number to the preceding term except the first term.

2) The fixed number is known as the common difference of an AP.

3) The common difference may be positive, negative, or zero.

4) If the first of an AP is "a" and the common difference is "d" then the terms of an

AP will be:  a, (a + d), (a + 2d), (a + 3d) ...

5) For the terms a₁, a₂, a₃, a_4, when a₂ - a₁ = a₃ - a₂ = a₄ - a₃ = d, then we say

that the terms a₁, a₂, a₃, a_4, are in AP.

Solution:

(i) 2, 4, 8, 16, . . . 

1) Here the first term is a = 2.

2) Here a₁ = a = 2, a₂ = 4, a₃ = 8, a₄ = 16.

3) Here, 

a) First difference:

d = a₂ - a₁ = 4 - 2

d = a₂ - a₁ = 2 --------- equation 1

b) Second difference:

d = a₃ - a₂ = 8 - 4

d = a₃ - a₂ = 4 --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ ≠ a₃ - a₂, so given terms are not in AP.

(ii) 2, 5/2, 3, 7/2, . . .

1) Here the first term is a = 2.

2) Here a₁ = a = 2, a₂ = 5/2, a₃ = 3, a₄ = 7/2.

3) Here, 

a) First difference:

d = a₂ - a₁ = 5/2 - 2

d = a₂ - a₁ = (5 - 4)/2 

d = a₂ - a₁ = 1/2 --------- equation 1

b) Second difference:

d = a₃ - a₂ = 3 - 5/2

d = a₃ - a₂ = (6 - 5)/2 

d = a₃ - a₂ = 1/2 --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ = a₃ - a₂, so given terms are in AP.

5) So, here the common difference d = 1/2.

6) So the next 3 terms are:

a) Fifth term:

a₅ =  a₄ + d

a₅ =  7/2 + 1/2

a₅ =  (7 + 1)/2

a₅ =  8/2

a₅ =  4

b) Sixth term:

a₆ =  a₅ + d

a₆ =  4 + 1/2

a₆ =  (8 + 1)/2

a₆ =  9/2

c) Seventh term:

a₇ =  a₆ + d

a₇ =  9/2 + 1/2

a₇ =  (9 + 1)/2

a₇ =  10/2

a₇ =  5

 7) The next 3 terms are 4, 9/2, and 5.

 

(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . .

1) Here the first term is a = - 1.2.

2) Here a₁ = a = - 1.2, a₂ = - 3.2, a₃ = - 5.2, a₄ = - 7.2.

3) Here, 

a) First difference:

d = a₂ - a₁ = (- 3.2) - (- 1.2)

d = a₂ - a₁ = - 3.2 + 1.2 

d = a₂ - a₁ = - 2 --------- equation 1

b) Second difference:

d = a₃ - a₂ = (- 5.2) - (- 3.2)

d = a₃ - a₂ = - 5.2 + 3.2 

d = a₃ - a₂ = - 2 --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ = a₃ - a₂, so given terms are in AP.

5) So, here the common difference is d = - 2.

6) So the next 3 terms are:

a) Fifth term:

a₅ =  a₄ + d

a₅ =  - 7.2 + (- 2)

a₅ =  - 7.2 - 2

a₅ =  - 9.2

b) Sixth term:

a₆ =  a₅ + d

a₆ =  - 9.2 + (- 2)

a₆ =  - 9.2 - 2

a₆ =  - 11.2

c) Seventh term:

a₇ =  a₆ + d

a₇ =  - 11.2 + (- 2)

a₇ =  - 11.2 - 2

a₇ =  - 13.2

 7) So the next 3 terms are - 9.2, - 11.2, and - 13.2.

(iv) – 10, – 6, – 2, 2, . . .

1) Here the first term is a = - 10.

2) Here a₁ = a = - 10, a₂ = - 6, a₃ = - 2, a₄ = 2.

3) Here, 

a) First difference:

d = a₂ - a₁ = (- 6) - (- 10)

d = a₂ - a₁ = - 6 + 10 

d = a₂ - a₁ = 4 --------- equation 1

b) Second difference:

d = a₃ - a₂ = (- 2) - (- 6)

d = a₃ - a₂ = - 2 + 6 

d = a₃ - a₂ = 4 --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ = a₃ - a₂, so given terms are in AP.

5) So, here the common difference is d = 4.

6) So the next 3 terms are:

a) Fifth term:

a₅ =  a₄ + d

a₅ =  2 + 4

a₅ =  6

b) Sixth term:

a₆ =  a₅ + d

a₆ =  6 + 2

a₆ =  8

c) Seventh term:

a₇ =  a₆ + d

a₇ =  8 + 2

a₇ =  10

 7) The next 3 terms are 6, 8, and 10.

(v) 3, 3 + √2 , 3 + 2√2 , 3 + 3 √2 , . . .

1) Here the first term is a = 3.

2) Here a₁ = a = 3, a₂ = 3 + √2, a₃ = 3 + 2√2, a₄ = 3 + 3√2.

3) Here, 

a) First difference:

d = a₂ - a₁ = (3 + √2) - (3)

d = a₂ - a₁ = 3 + √2 - 3 

d = a₂ - a₁ = √2 --------- equation 1

b) Second difference:

d = a₃ - a₂ = (3 + 2√2) - (3 + √2)

d = a₃ - a₂ = 3 + 2√2 - 3 - √2

d = a₃ - a₂ = √2 --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ = a₃ - a₂, so given terms are in AP.

5) So, here the common difference is d = √2.

6) So the next 3 terms are:

a) Fifth term:

a₅ =  a₄ + d

a₅ =  (3 + 3√2) + √2

a₅ =  3 + 4√2

b) Sixth term:

a₆ =  a₅ + d

a₆ =  (3 + 4√2) + √2

a₆ =  3 + 5√2

c) Seventh term:

a₇ =  a₆ + d

a₆ =  (3 + 5√2) + √2

a₆ =  3 + 6√2

 7) So next 3 terms are (3 + 4√2), (3 + 5√2), and (3 + 6√2).

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

1) Here the first term is a = 0.2.

2) Here a₁ = a = 0.2, a₂ = 0.22, a₃ = 0.222, a₄ = 0.2222.

3) Here, 

a) First difference:

d = a₂ - a₁ = (0.22) - (0.2)

d = a₂ - a₁ = 0.22 - 0.2 

d = a₂ - a₁ = 0.02 --------- equation 1

b) Second difference:

d = a₃ - a₂ = (0.222) - (0.22)

d = a₃ - a₂ = 0.222 - 0.22 

d = a₃ - a₂ = 0.002 --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ ≠ a₃ - a₂, so given terms are not in AP.

(vii) 0, – 4, – 8, –12, . . .

1) Here the first term is a = 0.

2) Here a₁ = a = 0, a₂ = - 4, a₃ = - 8, a₄ = - 12.

3) Here, 

a) First difference:

d = a₂ - a₁ = (- 4) - (0)

d = a₂ - a₁ = - 4 + 0 

d = a₂ - a₁ = - 4 --------- equation 1

b) Second difference:

d = a₃ - a₂ = (- 8) - (- 4)

d = a₃ - a₂ = - 8 + 4 

d = a₃ - a₂ = - 4 --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ = a₃ - a₂, so given terms are in AP.

5) So, here the common difference is d = - 4.

6) So the next 3 terms are:

a) Fifth term:

a₅ =  a₄ + d

a₅ =  - 12 + (- 4)

a₅ =  - 16

b) Sixth term:

a₆ =  a₅ + d

a₆ =  - 16 + (- 4)

a₆ =  - 20

c) Seventh term:

a₇ =  a₆ + d

a₇ =  - 20 + (- 4)

a₇ =  - 24

 7) The next 3 terms are - 16, - 20, and - 24.

(viii) - 1/2, - 1/2, - 1/2, - 1/2, . . .

1) Here the first term is a = - 1/2.

2) Here a₁ = a = - 1/2 , a₂ = - 1/2, a₃ = - 1/2, a₄ = - 1/2.

3) Here, 

a) First difference:

d = a₂ - a₁ = (- 1/2) - (- 1/2)

d = a₂ - a₁ = - 1/2 + 1/2 

d = a₂ - a₁ = 0 --------- equation 1

b) Second difference:

d = a₃ - a₂ = (- 1/2) - (- 1/2)

d = a₃ - a₂ = - 1/2 + 1/2  

d = a₃ - a₂ = 0 --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ = a₃ - a₂, so given terms are in AP.

5) So, here the common difference is d = 0.

6) So the next 3 terms are:

a) Fifth term:

a₅ =  a₄ + d

a₅ =  - 1/2 + 0

a₅ =  - 1/2

b) Sixth term:

a₆ =  a₅ + d

a₆ =  - 1/2 + 0

a₆ =  - 1/2

c) Seventh term:

a₇ =  a₆ + d

a₇ =  - 1/2 + 0

a₇ =  - 1/2

 7) So next 3 terms are - 1/2, - 1/2, and - 1/2.

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

1) Here the first term is a = 1.

2) Here a₁ = a = 1 , a₂ = 3 a₃ = 9, a₄ = 27.

3) Here, 

a) First difference:

d = a₂ - a₁ = (3) - (1)

d = a₂ - a₁ = 3 - 1 

d = a₂ - a₁ = 2 --------- equation 1

b) Second difference:

d = a₃ - a₂ = 9 - 3

d = a₃ - a₂ = 6 --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ ≠ a₃ - a₂, so given terms are not in AP.

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

1) Here the first term is a = a.

2) Here a₁ = a = a, a₂ = 2a, a₃ = 3a, a₄ = 4a.

3) Here, 

a) First difference:

d = a₂ - a₁ = (2a) - (a)

d = a₂ - a₁ = 2a - a

d = a₂ - a₁ = a --------- equation 1

b) Second difference:

d = a₃ - a₂ = (3a) - (2a)

d = a₃ - a₂ = 3a - 2a 

d = a₃ - a₂ = a --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ = a₃ - a₂, so given terms are in AP.

5) So, here common difference d = a.

6) So the next 3 terms are:

a) Fifth term:

a₅ =  a₄ + d

a₅ =  4a + a

a₅ =  5a

b) Sixth term:

a₆ =  a₅ + d

a₆ =  5a + a

a₆ =  6a

c) Seventh term:

a₇ =  a₆ + d

a₇ =  6a + a

a₇ =  7a

 7) The next 3 terms are 5a, 6a, and 7a.

(xi) a, a², a³, a⁴, . . .

1) Here the first term is a = a.

2) Here a₁ = a = a, a₂ = a², a₃ = a³, a₄ = a⁴.

3) Here, 

a) First difference:

d = a₂ - a₁ = (a²) - (a)

d = a₂ - a₁ = a(a - 1) --------- equation 1

b) Second difference:

d = a₃ - a₂ = (a³) - (a²)

d = a₃ - a₂ = a²(a - 1) --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ ≠ a₃ - a₂, so given terms are not in AP.

(xii) √2, √8, √18 , √32, . . .

1) Here the first term is a = √2.

2) Here a₁ = a = √2, a₂ = √8, a₃ = √18, a₄ = √32.

3) Here, 

a) First difference:

d = a₂ - a₁ = √8 - √2

d = a₂ - a₁ = 2√2 - √2

d = a₂ - a₁ = √2 --------- equation 1

b) Second difference:

d = a₃ - a₂ = √18 - √8

d = a₃ - a₂ = 3√2 - 2√2

d = a₃ - a₂ = √2 --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ = a₃ - a₂, so given terms are in AP.

5) So, here the common difference is d = √2.

6) So the next 3 terms are:

a) Fifth term:

a₅ =  a₄ + d

a₅ =  √32 + √2

a₅ =  4√2 + √2

a₅ =  5√2

a₅ =  √50 

b) Sixth term:

a₆ =  a₄ + d
a₆ =  √50 + √2

a₆ =  5√2 + √2

a₆ =  6√2
a₆ =  √72 

c) Seventh term:

a₇ =  a₄ + d
a₇ =  √72 + √2

a₇ =  6√2 + √2

a₇ =  7√2
a₇ =  √98 

 7) The next 3 terms are √50, √72, and √98.

(xiii) √3, √6, √9 , √12 , . . .

1) Here the first term is a = √3.

2) Here a₁ = a = √3, a₂ = √6, a₃ = √9, a₄ = √12.

3) Here, 

a) First difference:

d = a₂ - a₁ = √6 - √3

d = a₂ - a₁ = √3√2 - √3

d = a₂ - a₁ = √3(√2 - 1) --------- equation 1

b) Second difference:

d = a₃ - a₂ = √9 - √6

d = a₃ - a₂ = √3√3 - √3√2

d = a₃ - a₂ = √3(√3 - √2) --------- equation 2 

4) From equation 1 and equation 2, a₂ - a₁ ≠ a₃ - a₂, so given terms are not in AP.

(xiv) 1², 3², 5², 7², . . .

1) Here the first term is a = 1².

2) Here a₁ = a = 1², a₂ = 3², a₃ = 5³, a₄ = 7⁴.

3) Here, 

a) First difference:

d = a₂ - a₁ = 3² - 1²

d = a₂ - a₁ = 9 - 1 

d = a₂ - a₁ = 8 --------- equation 1

b) Second difference:

d = a₃ - a₂ = 5² - 3²

d = a₃ - a₂ = 25 - 9 

d = a₃ - a₂ = 16 --------- equation 2

4) From equation 1 and equation 2, a₂ - a₁ ≠ a₃ - a₂, so given terms are not in AP.

(xv) 1², 5², 7², 7³, . . .

1) Here the first term is a = 1².

2) Here a₁ = a = 1², a₂ = 5², a₃ = 7², a₄ = 73.

3) Here, 

a) First difference:

d = a₂ - a₁ = 5² - 1²

d = a₂ - a₁ = 25 - 1 

d = a₂ - a₁ = 24 --------- equation 1

b) Second difference:

d = a₃ - a₂ = 7² - 5²

d = a₃ - a₂ = 49 - 25 

d = a₃ - a₂ = 24 --------- equation 2

c) Third difference:

d = a₄ - a₃ = 73 - 7²

d = a₄ - a₃ = 73 - 49

d = a₄ - a₃ = 24 --------- equation 3

4) From equation 1, 2 and 3, a₂ - a₁ = a₃ - a₂ = a₄ - a₃, so given terms are in AP.

5) So, here the common difference is d = 24.

6) So the next 3 terms are:

a) Fifth term:

a₅ = a₄ + d

a₅ = 73 + 24

a₅ = 97 

b) Sixth term:

a₆ =  a₄ + d
a₆ = 97 + 24

a₆ = 121 

c) Seventh term:

a₇ =  a₄ + d
a₇ = 121 + 24

a₇ = 145 

 7) The next 3 terms are 97, 121, and 145.

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