**(1) a = 10, d = 5**

**Solution:**

Given a = 10, d = 5

Let a_{1} = a = 10

Since, the common difference d = 5

Using formula a_{n + 1} = a_{n} + d

Thus, a_{2} = a_{1} + d = 10 + 5 = 15

a_{3 =} a_{2} + d = 15 + 5 = 20

a_{4} = a_{3} + 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 a_{1} = a = – 3

Since, the common difference d = 0

Using formula a_{n + 1} = a_{n} + d

Thus, a_{2} = a_{1} + d = – 3 + 0 = – 3

a_{3 =} a_{2} + d = – 3 + 0 = – 3

a_{4} = a_{3} + 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 a_{1} = a = – 1.25

Since, the common difference d = 3

Using formula a_{n + 1} = a_{n} + d

Thus, a_{2} = a_{1} + d = – 1.25 + 3 = 1.75

a_{3 =} a_{2} + d = 1.75 + 3 = 4.75

a_{4} = a_{3} + 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 a_{1} = a = 6

Since, the common difference d = – 3

Using formula a_{n + 1} = a_{n} + d

Thus, a_{2} = a_{1} + d = 6 + (– 3) = 6 – 3 = 3

a_{3 =} a_{2} + d = 3 + (– 3) = 3 – 3 = 0

a_{4} = a_{3} + 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 a_{1} = a = – 19

Since, the common difference d = – 4

Using formula a_{n + 1} = a_{n} + d

Thus, a_{2} = a_{1} + d = – 19 + (– 4) = – 19 – 4 = – 23

a_{3 =} a_{2} + d = – 23 + (– 4) = – 23 – 4 = – 27

a_{4} = a_{3} + d = – 27 + (– 4) = – 27 – 4 = – 31

Hence, An A.P with common difference – 4 is – 19, – 23, – 27, – 31, ….