Given: a1 = a = 5 ….. (1)
a4 = a + 3d = 9\(\frac{1}{2}\) ….. (2)
Solving equations (1) and (2);
⇒ 3d = 4\(\frac{1}{2}\)
⇒ 3d = \(\frac{9}{2}\)
⇒ d = \(\frac{9}{2\times3}\) = \(\frac{3}{2}\)
∴ a2 = a + d = 5 + \(\frac{3}{2}\) = \(\frac{13}{2}\)
a3 = a + 2d = 5 + 2 × \(\frac{3}{2}\) = 5 + 3 = 8