(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = 1/2
(v) a = -1.25, d = -0.25
Solution:
(i) a = 10, d = 10
Let us consider, the AP as 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 …
The first four terms of this A.P. will be 10, 20, 30, and 40.
(ii) a = – 2, d = 0
Let us consider, the AP as 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 …
The first four terms of this A.P. will be – 2, – 2, – 2 and – 2.
(iii) a = 4, d = – 3
Let us consider, the AP as 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 …
The first four terms of this A.P. will be 4, 1, – 2, and – 5.
(iv) a = – 1, d = 1/2
Let us consider, the AP as 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
The first four terms of this A.P. will be -1, -1/2, 0, and 1/2.
(v) a = – 1.25, d = – 0.25
Let us consider, the AP as 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 ……..
The first four terms of this A.P. will be – 1.25, – 1.50, – 1.75, and – 2.00.