Write first four terms of the AP, when the first term \(a\) and the common difference \(d\) are given as follows:
(i) \(a = 10,\ d = 10\)
(ii) \(a = -2,\ d = 0\)
(iii) \(a = 4,\ d = -3\)
(iv) \(a = -1,\ d = \dfrac{1}{2}\)
(v) \(a = -1.25,\ d = 0.25\)
(i) \(10, 20, 30, 40\)
(ii) \(-2, -2, -2, -2\)
(iii) \(4, 1, -2, -5\)
(iv) \(-1, -\tfrac{1}{2}, 0, \tfrac{1}{2}\)
(v) \(-1.25, -1.50, -1.75, -2.00\)
Recall: In an arithmetic progression (AP) with first term \(a\) and common difference \(d\), the terms are:
\[a,\ a + d,\ a + 2d,\ a + 3d,\dots\]
So, the first four terms are: \(T_1 = a\), \(T_2 = a + d\), \(T_3 = a + 2d\), \(T_4 = a + 3d\).
\(T_1 = 10\)
\(T_2 = 10 + 10 = 20\)
\(T_3 = 10 + 2 \times 10 = 30\)
\(T_4 = 10 + 3 \times 10 = 40\)
First four terms: \(10, 20, 30, 40\).
Here the common difference is 0, so every term is the same.
\(T_1 = -2\)
\(T_2 = -2 + 0 = -2\)
\(T_3 = -2 + 2 \times 0 = -2\)
\(T_4 = -2 + 3 \times 0 = -2\)
First four terms: \(-2, -2, -2, -2\).
\(T_1 = 4\)
\(T_2 = 4 + (-3) = 1\)
\(T_3 = 4 + 2(-3) = 4 - 6 = -2\)
\(T_4 = 4 + 3(-3) = 4 - 9 = -5\)
First four terms: \(4, 1, -2, -5\).
\(T_1 = -1\)
\(T_2 = -1 + \dfrac{1}{2} = -\dfrac{1}{2}\)
\(T_3 = -1 + 2 \times \dfrac{1}{2} = -1 + 1 = 0\)
\(T_4 = -1 + 3 \times \dfrac{1}{2} = -1 + \dfrac{3}{2} = \dfrac{1}{2}\)
First four terms: \(-1, -\tfrac{1}{2}, 0, \tfrac{1}{2}\).
Starting from \(-1.25\), each successive term changes by \(0.25\) in the same direction, giving:
\(T_1 = -1.25\)
\(T_2 = -1.50\)
\(T_3 = -1.75\)
\(T_4 = -2.00\)
First four terms: \(-1.25, -1.50, -1.75, -2.00\).