Write first four terms of the AP, when the first term a and common difference d are given as follows:
(i)a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = \({{1} \over {2}}\)
(v) a = -1.25, d = -0.25

(i) First term = a = 10, d = 10
Second term = a + d = 10 + 10 = 20
Third term = second term + d = 20 + 10 = 30
Fourth term = third term + d = 30 + 10 = 40
Therefore, first four terms are: 10, 20, 30, 40

(ii) First term = a = –2 , d = 0
Second term = a + d = –2 + 0 = –2
Third term = second term + d = –2 + 0 = –2
Fourth term = third term + d = –2 + 0 = –2
Therefore, first four terms are: –2, –2, –2, –2

(iii) First term = a = 4, d =–3
Second term = a + d = 4 – 3 = 1
Third term = second term + d = 1 – 3 = –2
Fourth term = third term + d = –2 – 3 = –5
Therefore, first four terms are: 4, 1, –2, –5

(iv) First term = a = –1, d = \({{1} \over {2}}\)
Second term = a + d = –1 + \({{1} \over {2}}\) = \({{-1} \over {2}}\)
Third term = second term + d = \({{-1} \over {2}} + {{1} \over {2}}\) = 0
Fourth term = third term + d = 0 + \({{1} \over {2}}\) = \({{1} \over {2}}\)
Therefore, first four terms are: –1, \({{-1} \over {2}}\), 0, \({{1} \over {2}}\)

(v) First term = a = –1.25, d = –0.25
Second term = a + d = –1.25 – 0.25 = –1.50
Third term = second term + d = –1.50 – 0.25 = –1.75
Fourth term = third term + d = –1.75 – 0.25 = –2.00
Therefore, first four terms are: –1.25, –1.50, –1.75, –2.00