Write the correct number in the given boxes from the following A. P.

(1) 1, 8, 15, 22, . . .

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 = t_{n + 1} – t_n
Thus, t₂ – t₁ = 15 – 8 = 7
t₃ – t₂ = 22 – 15 = 7
Thus, d = 7

(2) 3, 6, 9, 12, . . .

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 = t_{n + 1} – t_n
Thus, t₂ – t₁ = 9 – 6 = 3
t₃ – t₂ = 12 – 9 = 3
Thus, d = 3

(3) – 3, – 8, – 13, – 18, . . .

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 = t_{n + 1} – t_n
Thus, t₂ – t₁ = – 13 – (– 8) = – 13 + 8 = – 5
t₃ – t₂ = – 18 – (– 13) = – 18 + 13 = – 5
Thus, d = – 5

(4) 70, 60, 50, 40, . . .

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 = t_{n + 1} – t_n
Thus, t₂ – t₁ = 50 – 60 = – 10
t₃ – t₂ = 40 – 50 = – 10
Thus, d = – 10