(a) The third term of a Geometric Progression (G.P) is 24 and its seventh term is \(4\frac{20}{27}\). Find its first term.
(b) Given that y varies directly as x and inversely as the square of z. If y = 4, when x = 3 and z = 1, find y when x = 3 and z = 2.
Explanation
\(T_{n} = ar^{n - 1}\) (Terms of a G.P)
\(T_{3} = ar^{2} =24 .... (1)\)
\(T_{7} = ar^{6} = 4\frac{20}{27} = \frac{128}{27} .......... (2)\)
Divide (2) by (1),
\(\frac{ar^{6}}{ar^{2}} = \frac{\frac{128}{27}}{24}\)
\(r^{4} = \frac{16}{81}\)
\(r = \sqrt[4]{\frac{16}{81}}\)
= \(\frac{2}{3}\)
Putting r in equation (1),
\(ar^{2} = a(\frac{2}{3})^{2} = 24\)
\(\frac{4}{9} a = 24 \implies a = \frac{24 \times 9}{4}\)
a = 54.
(b) \(y \propto x \propto \frac{1}{z^{2}}\)
\(\implies y = \frac{kx}{z^{2}}\)
\(4 = \frac{3k}{1^{2}} \implies k = \frac{4}{3}\)
\(\therefore y = \frac{4x}{3z^{2}}\)
When x = 3, z = 2, y = ?
\(y = \frac{4(3)}{3(2^{2})}\)
= \(\frac{12}{12} = 1\).