Substituting n = 1, 2, 3, 4, 5, we obtain
a1 = 2×1-\(\frac{3}{6} =\frac{-1}{6}\)
a2 = 2×2-\(\frac{3}{6} = \frac{1}{6}\)
a3 = 2×3-\(\frac{3}{6} = \frac{3}{6} = \frac{1}{2}\)
a4 = 2×4-\(\frac{3}{6} = \frac{5}{6}\)
a5 = 2×5-\(\frac{3}{6} = \frac{7}{6}\)
Therefore, the required terms are \(\frac{-1}{6}, \frac{1}{6},\frac{1}{2}, \frac{5}{6}, and \frac{7}{6}\)
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?