(a) What is the 25th term of 5, 9, 13,... ? (b) Find the 5th

Explanation

(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 \(108, 72, 48, 32\).