SECTION B
Given that (x + 2), (4x + 3), and (7x + 24) are consecutive terms of a Geometric Progression (G.P.), find the:
a) value of x
b) common ratio.
Explanation
The term of the given G.P. are (x + 2), (4x + 3), and (7x + 24)
(a) Common ratio = \(\frac{(4x + 3)}{(x + 2)}\) = \(\frac{(7x + 24)}{(4x + 3)}\)
cross multiply
(4x + 3)(4x + 3) = (x + 2)(7x + 24)
16x\(^2\) + 24x + 9 = 7x\(^2\) + 38x + 48
16x\(^2\) - 7x\(^2\) + 24x - 38x + 9 - 48 = 0
9x\(^2\) - 14x - 39 = 0
9x\(^2\) - 27x + 13x - 39 = 0 (from 9x\(^2\) x 39)
9x(x - 3) + 13(x -3) = 0
(9x + 13)(x - 3) = 0
(9x + 13) = 0 or (x - 3) = 0
x = \(\frac{-13}{9}\) or 3
(b) When x = \(\frac{-13}{9}\)
Common ratio = \(\frac{(4x + 3)}{(x + 2)}\) = \(\frac{(4(\frac{-13}{9}) + 3)}{((\frac{-13}{9}) + 2)}\)
= \(\frac{\frac{-25}{9}}{\frac{5}{9}}\) = \(\frac{- 25}{9}\) x \(\frac{9}{5}\) = - 5.
Common ratio,r = - 5
When x = 3,
Common ratio = \(\frac{(4(3) + 3)}{((3) + 2)}\) = \(\frac{15}{5}\) = 3.
Common ratio, r = 3