(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 \(48\) and \(14\frac{2}{9}\) respectively. Write down the first four terms of the G.P.
(a) 5, 9, 13, ... is an A.P with the first term = 5 and common difference = 9 - 5 = 13 - 9 = 4.
\(T_{n} = a + (n - 1)d\) (terms of an A.P)
\(T_{25} = 5 + (25 - 1) \times 4\)
= \(5 + 24 \times 4\)
= \( 5 + 96 = 101\)
(b) \(\frac{8}{9}, \frac{-4}{3}, 2, ...\) is a G.P with first term = \(\frac{8}{9}\)
\(r = \frac{T_{2}}{T_{1}} = -\frac{4}{3} \div \frac{8}{9} = \frac{-4}{3} \times \frac{9}{8} = -\frac{3}{2}\)
\(T_{n} = ar^{n - 1}\) (terms of a G.P)
\(T_{5} = (\frac{8}{9})(-\frac{3}{2})^{5 - 1}\)
= \(\frac{8}{9} \times \frac{81}{16} = \frac{9}{2}\)
(c) \(T_{n} = ar^{n - 1}\) (terms of a G.P)
\(T_{3} = ar^{2} = 48 ... (1)\)
\(T_{6} = ar^{5} = 14\frac{2}{9} ... (2)\)
\((2) \div (1) : r^{3} = \frac{128}{9} \div 48 = \frac{128}{9} \times \frac{1}{48}\)
\(r^{3} = \frac{8}{27}\)
\(r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}\)
From (1), \(ar^{2} = 48 \implies a \times (\frac{2}{3})^{2} = 48\)
\(4a = 48 \times 9 \implies a = \frac{48 \times 9}{4} = 108\)
\(T_{2} = 108 \times \frac{2}{3} = 72\)
\(T_{4} = 48 \times \frac{2}{3} = 32\)
\(\therefore\) The first four terms of the sequence are \(108, 72, 48, 32\).
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