Write an A.P. whose first term is a and common difference is d in each of the following.

(1) a = 10, d = 5

Solution:
Given a = 10, d = 5
Let a₁ = a = 10
Since, the common difference d = 5
Using formula a_{n + 1} = a_n + d
Thus, a₂ = a₁ + d = 10 + 5 = 15
a_{3 =} a₂ + d = 15 + 5 = 20
a₄ = a₃ + d = 20 + 5 = 25
Hence, An A.P with common difference 5 is 10, 15, 20, 25, ….

(2) a = – 3, d = 0

Solution:
Given a = – 3, d = 0
Let a₁ = a = – 3
Since, the common difference d = 0
Using formula a_{n + 1} = a_n + d
Thus, a₂ = a₁ + d = – 3 + 0 = – 3
a_{3 =} a₂ + d = – 3 + 0 = – 3
a₄ = a₃ + d = – 3 + 0 = – 3
Hence, An A.P with common difference 0 is – 3, – 3, – 3, – 3, ….

(3) a = – 7, d = ½

Solution:

(4) a = – 1.25, d = 3

Solution:
Given a = – 1.25, d = 3
Let a₁ = a = – 1.25
Since, the common difference d = 3
Using formula a_{n + 1} = a_n + d
Thus, a₂ = a₁ + d = – 1.25 + 3 = 1.75
a_{3 =} a₂ + d = 1.75 + 3 = 4.75
a₄ = a₃ + d = 4.75 + 3 = 7.75
Hence, An A.P with common difference 3 is – 1.25, 1.75, 4.75, 7.75

(5) a = 6, d = – 3

Solution:
Given a = 6, d = – 3
Let a₁ = a = 6
Since, the common difference d = – 3
Using formula a_{n + 1} = a_n + d
Thus, a₂ = a₁ + d = 6 + (– 3) = 6 – 3 = 3
a_{3 =} a₂ + d = 3 + (– 3) = 3 – 3 = 0
a₄ = a₃ + d = 0 + (– 3) = – 3
Hence, An A.P with common difference – 3 is 6, 3, 0, – 3…

(6) a = – 19, d = – 4

Solution:
Given a = – 19, d = – 4
Let a₁ = a = – 19
Since, the common difference d = – 4
Using formula a_{n + 1} = a_n + d
Thus, a₂ = a₁ + d = – 19 + (– 4) = – 19 – 4 = – 23
a_{3 =} a₂ + d = – 23 + (– 4) = – 23 – 4 = – 27
a₄ = a₃ + d = – 27 + (– 4) = – 27 – 4 = – 31
Hence, An A.P with common difference – 4 is – 19, – 23, – 27, – 31, ….