(1) 1, 8, 15, 22, . . .
![Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter3 Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 6](https://cdn1.byjus.com/wp-content/uploads/2020/04/maharashtra-board-solutions-for-class-10-maths-part-1-chapter-2-image-610.png)
Solution:
Given 1, 8, 15, 22, . . .
First term a = 1
Second term t1 = 8
Third term t2 = 15
Fourth term t3 = 22
We know that d = tn + 1 – tn
Thus, t2 – t1 = 15 – 8 = 7
t3 – t2 = 22 – 15 = 7
Thus, d = 7
(2) 3, 6, 9, 12, . . .
![Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter3 Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 7](https://cdn1.byjus.com/wp-content/uploads/2020/04/maharashtra-board-solutions-for-class-10-maths-part-1-chapter-2-image-710.png)
Solution:
Given 3,6,9,12, . . .
First term a = 3
Second term t1 = 6
Third term t2 = 9
Fourth term t3 = 12
We know that d = tn + 1 – tn
Thus, t2 – t1 = 9 – 6 = 3
t3 – t2 = 12 – 9 = 3
Thus, d = 3
(3) – 3, – 8, – 13, – 18, . . .
![Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter3 Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 8](https://cdn1.byjus.com/wp-content/uploads/2020/04/maharashtra-board-solutions-for-class-10-maths-part-1-chapter-2-image-810.png)
Solution:
Given – 3, – 8, – 13, – 18, . . .
First term a = – 3
Second term t1 = – 8
Third term t2 = – 13
Fourth term t3 = – 18
We know that d = tn + 1 – tn
Thus, t2 – t1 = – 13 – (– 8) = – 13 + 8 = – 5
t3 – t2 = – 18 – (– 13) = – 18 + 13 = – 5
Thus, d = – 5
(4) 70, 60, 50, 40, . . .
![Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter3 Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 9](https://cdn1.byjus.com/wp-content/uploads/2020/04/maharashtra-board-solutions-for-class-10-maths-part-1-chapter-2-image-910.png)
Solution:
Given 70, 60, 50, 40, . . .
First term a = 70
Second term t1 = 60
Third term t2 = 50
Fourth term t3 = 40
We know that d = tn + 1 – tn
Thus, t2 – t1 = 50 – 60 = – 10
t3 – t2 = 40 – 50 = – 10
Thus, d = – 10