(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 ...
Question 1 Report
(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.
Answer Details
(a) For a G.P. with first term \(a\) and common ratio \(r\):
Third term: \(ar^2 = 24\) ... (1)
Seventh term: \(ar^6 = 4\tfrac{20}{27} = \dfrac{128}{27}\) ... (2)
Divide (2) by (1):
\[r^4 = \frac{128/27}{24} = \frac{128}{648} = \frac{16}{81}.\]
\[r^2 = \sqrt{\frac{16}{81}} = \frac{4}{9} \Rightarrow r = \frac{2}{3}.\]
From (1): \(a\left(\dfrac{4}{9}\right) = 24 \Rightarrow a = 24 \times \dfrac{9}{4} = 54\).
The first term is \(a = \mathbf{54}\).
(b) \(y\) varies directly as \(x\) and inversely as the square of \(z\):
\[y = \frac{kx}{z^2}.\]
When \(y = 4, x = 3, z = 1\): \(4 = \dfrac{k(3)}{1^2} \Rightarrow k = \dfrac{4}{3}\).
When \(x = 3, z = 2\):
\[y = \frac{\tfrac{4}{3}(3)}{2^2} = \frac{4}{4} = 1.\]
Therefore \(y = \mathbf{1}\).
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