(a) What is the 25th term of 5, 9, 13,... ? (b) Find the 5th term of \(\frac{8}{9}, \frac{-4}{3}, 2, ...\). (c) The 3rd and 6th terms of a G.P are \(48\) an...

Answer Details

(a) A.P. \(5, 9, 13, \dots\) with \(a = 5,\; d = 4\).

\[T_{25} = a + 24d = 5 + 24(4) = 5 + 96 = 101\]

(b) G.P. \(\dfrac{8}{9},\, -\dfrac{4}{3},\, 2, \dots\) with \(a = \dfrac{8}{9}\) and common ratio

\[r = \frac{-4/3}{8/9} = -\frac{4}{3}\times\frac{9}{8} = -\frac{3}{2}\]

\[T_5 = ar^4 = \frac{8}{9}\left(-\frac{3}{2}\right)^4 = \frac{8}{9}\times\frac{81}{16} = \frac{9}{2} = 4\tfrac12\]

(c) For the G.P., \(T_3 = ar^2 = 48\) and \(T_6 = ar^5 = 14\tfrac29 = \dfrac{128}{9}\). Dividing:

\[r^3 = \frac{ar^5}{ar^2} = \frac{128/9}{48} = \frac{8}{27} \;\Rightarrow\; r = \frac{2}{3}\]

\[ar^2 = a\left(\tfrac49\right) = 48 \;\Rightarrow\; a = 108\]

The first four terms are \(108,\; 72,\; 48,\; 32\).