Solution
(i) a = 10, d =10
Let us consider, the Arithmetic Progressionseries be a1, a2, a3, a4, a5 …
a1 = a = 10
a2 = a1+d = 10+10 = 20
a3 = a2+d = 20+10 = 30
a4 = a3+d = 30+10 = 40
a5 = a4+d = 40+10 = 50
And so on…
Therefore, the A.P. series will be 10, 20, 30,40, 50 …
And First four terms of this A.P. will be 10,20, 30, and 40.
Solution
(ii) a = – 2, d =0
Let us consider, the Arithmetic Progressionseries be a1, a2, a3, a4, a5 …
a1 = a = -2
a2 = a1+d = – 2+0 = – 2
a3 = a2+d = – 2+0 = – 2
a4 = a3+d = – 2+0 = – 2
Therefore, the A.P. series will be – 2, – 2, –2, – 2 …
And, First four terms of this A.P. will be –2, – 2, – 2 and – 2.
Solution
(iii) a = 4, d =– 3
Let us consider, the Arithmetic Progressionseries be a1, a2, a3, a4, a5 …
a1 = a = 4
a2 = a1+d = 4-3 = 1
a3 = a2+d = 1-3 = – 2
a4 = a3+d = -2-3 = – 5
Therefore, the A.P. series will be 4, 1, – 2 –5 …
And, first four terms of this A.P. will be 4,1, – 2 and – 5.
Solution
(iv) a = – 1, d =1/2
Let us consider, the Arithmetic Progressionseries be a1, a2, a3, a4, a5 …
a2 = a1+d = -1+1/2 = -1/2
a3 = a2+d = -1/2+1/2 = 0
a4 = a3+d = 0+1/2 = 1/2
Thus, the A.P. series will be-1, -1/2, 0, 1/2
And First four terms of this A.P. will be -1,-1/2, 0 and 1/2.
Solution
(v) a = – 1.25, d =– 0.25
Let us consider, the Arithmetic Progressionseries be a1, a2, a3, a4, a5 …
a1 = a = – 1.25
a2 = a1 + d = –1.25-0.25 = – 1.50
a3 = a2 + d = –1.50-0.25 = – 1.75
a4 = a3 + d = –1.75-0.25 = – 2.00
Therefore, the A.P series will be 1.25, –1.50, – 1.75, – 2.00 ……..
And first four terms of this A.P. will be –1.25, – 1.50, – 1.75 and – 2.00.