Write the first three terms of the AP given \(a\) and \(d\).
(i) \(a=\dfrac{1}{2},\; d=-\dfrac{1}{6}\)
(ii) \(a=-5,\; d=-3\)
(iii) \(a=\sqrt{2},\; d=\dfrac{1}{\sqrt{2}}\)
(i) \(\dfrac{1}{2},\; \dfrac{1}{3},\; \dfrac{1}{6}\)
(ii) \(-5,\; -8,\; -11\)
(iii) \(\sqrt{2},\; \sqrt{2}+\dfrac{1}{\sqrt{2}},\; \sqrt{2}+\dfrac{2}{\sqrt{2}}=2\sqrt{2}\)
An Arithmetic Progression (AP) is a list of numbers where each term is obtained by adding the common difference (\(d\)) to the previous term.
The first three terms are always:
\(T_1 = a\) (the first term)
\(T_2 = a + d\) (first term + common difference)
\(T_3 = a + 2d\) (first term + twice the common difference)
(i) \(a = \tfrac{1}{2}, d = -\tfrac{1}{6}\)
\(T_1 = a = \tfrac{1}{2}\)
\(T_2 = a + d = \tfrac{1}{2} - \tfrac{1}{6} = \tfrac{3}{6} - \tfrac{1}{6} = \tfrac{2}{6} = \tfrac{1}{3}\)
\(T_3 = a + 2d = \tfrac{1}{2} + 2\left(-\tfrac{1}{6}\right) = \tfrac{1}{2} - \tfrac{2}{6} = \tfrac{3}{6} - \tfrac{2}{6} = \tfrac{1}{6}\)
(ii) \(a = -5, d = -3\)
\(T_1 = a = -5\)
\(T_2 = a + d = -5 + (-3) = -8\)
\(T_3 = a + 2d = -5 + 2(-3) = -5 - 6 = -11\)
(iii) \(a = \sqrt{2}, d = \tfrac{1}{\sqrt{2}}\)
\(T_1 = a = \sqrt{2}\)
\(T_2 = a + d = \sqrt{2} + \tfrac{1}{\sqrt{2}}\)
\(T_3 = a + 2d = \sqrt{2} + 2\left(\tfrac{1}{\sqrt{2}}\right) = \sqrt{2} + \tfrac{2}{\sqrt{2}} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}\)
So, the first three terms of each AP are as written in the answer.