198-NCERT New Syllabus Grade 10 Arithmetic Progressions Ex-5.1

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NCERT New Syllabus Mathematics

Class: 10

Exercise 5.1

Topic: Arithmetic Progressions

Introduction to Arithmetic Progressions

Arithmetic Progressions (AP) are a fundamental mathematical concept that forms the basis for understanding various real-world applications, ranging from financial planning to engineering design. In this chapter of the 10th-grade NCERT syllabus, we delve into sequences where the difference between consecutive terms remains constant. This simple yet powerful concept helps students recognize patterns and solve problems involving future predictions, geometric designs, and even natural phenomena.

An arithmetic progression represented as a sequence of numbers like a, a+d, a+2d, and so on, is defined by the first term (a) and the common difference (d). This progression opens up a range of possibilities for solving complex problems by understanding the behavior of numbers in a linear format.

In this blog, we’ll explore the key concepts of AP, including its general form, the formula for the nth term, and the sum of n terms. We’ll also solve a variety of problems from the new NCERT syllabus, making it easier for students to grasp the concept and excel in their exams.

Explanation:

An arithmetic progression (AP) is a sequence of numbers where each term, after the first, is generated by adding a constant value, known as the common difference, to the preceding term. This common difference can be positive, negative, or zero, depending on the nature of the progression.

In an arithmetic progression (AP), the terms follow a predictable pattern. If the first term of the sequence is denoted by a, and the common difference is represented by $d$, then the terms of the AP can be written as:

a, a + d, a + 2d, a + 3d, . . .

This sequence is known as the general form of an arithmetic progression. Each successive term is obtained by adding the common difference $d$ to the preceding term.

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.

3) Yes. 

The sequence 15, 23, 31, ... forms an Arithmetic Progression (AP) because each succeeding term is obtained by adding 8 to the preceding term.

(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.

150, 200, 250, . . . form 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, 

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, and d = 1/2.

 

(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) The next 3 terms are – 9.2, – 11.2, and – 13.2, d = – 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 + 4

a₆ =  10

c) Seventh term:

a₇ =  a₆ + d

a₇ =  10 + 4

a₇ =  14

 7) The next 3 terms are 6, 10, and 14, d = 4.

(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) The next 3 terms are (3 + 4√2), (3 + 5√2), and (3 + 6√2), d = √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, 

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, d = – 4.

(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, d = 0.

(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, 

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, d = a.

(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, 

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, d = √2.

(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, 

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.

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, d = 24.

Conclusion: The Power of Arithmetic Progressions

As we wrap up our journey through Arithmetic Progressions, we've seen how sequences unfold and reveal patterns in everyday life—from predicting future events to organizing data. The concept of common differences and the sum of terms allows us to model various scenarios in math and real-world situations. Whether you’re solving problems in exams or exploring the magic of number sequences, AP knowledge is a powerful tool. Keep practicing; you’ll see how every sequence leads to new possibilities!

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#ArithmeticProgressions #Class10Math #MathematicsNCERT #APFormulas #MathHelp #LearnMath #MathConcepts #CBSEMath #ProgressionPatterns #MathSequences #MathInLife #NCERTClass10 #SequenceAndPattern #MathWizards #LearningWithNumbers #MathMastery #EndlessPossibilities

Click here to explore the next step ⇨ NCERT New Syllabus Class 10 - Arithmetic Progressions Exercise 5.2

Anil Satpute