(a) The third term of a Geometric Progression (G.P) is 24 and its seventh term is (4frac{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} = 4frac{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).