Write first four of the terms of the A.P., when the first term a and the common difference a are given as follows :

(i) a = 10, d = 10

(ii) a = -2, d = 0

(iii) a = 4, d = -3

(iv) a = $\dfrac{1}{2}$, d = $-\dfrac{1}{6}$

(i) Here, a₁ = a = 10,   a₂ = a₁ + d = 10 + 10 = 20,

a₃ = a₂ + d = 20 + 10 = 30,    a₄ = a₃ + d = 30 + 10 = 40.

Hence, the first four terms of A.P. are 10, 20, 30, 40.

(ii) Here, a₁ = a = -2,   a₂ = a₁ + d = -2 + 0 = -2,

a₃ = a₂ + d = -2 + 0 = -2,    a₄ = a₃ + d = -2 + 0 = -2.

Hence, the first four terms of A.P. are -2, -2, -2, -2.

(iii) Here, a₁ = a = 4,   a₂ = a₁ + d = 4 + (-3) = 1,

a₃ = a₂ + d = 1 + (-3) = -2,    a₄ = a₃ + d = -2 + (-3) = -5.

Hence, the first four terms of A.P. are 4, 1, -2, -5.

(iv) Here, a₁ = a = $\dfrac{1}{2}$,   a₂ = a₁ + d = $\dfrac{1}{2} + \big(-\dfrac{1}{6}\big) = \dfrac{2}{6} = \dfrac{1}{3}$,

a₃ = a₂ + d = $\dfrac{2}{6} + \big(-\dfrac{1}{6}\big) = \dfrac{1}{6}$,    a₄ = a₃ + d = $\dfrac{1}{6} + \big(-\dfrac{1}{6}\big)$ = 0.

Hence, the first four terms of A.P. are $\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{6}, 0$.