9709 P12 - Nov 2019 - Q8
888
The first, second and third terms of a geometric progression are \(x\), \(x - 3\) and \(x - 5\) respectively.
- Find the value of \(x\).
- Find the fourth term of the progression.
- Find the sum to infinity of the progression.
Solution
(i) For a geometric progression, the ratio between consecutive terms is constant. Therefore, \(\frac{x-3}{x} = \frac{x-5}{x-3}\).
Cross-multiplying gives: \((x-3)^2 = x(x-5)\).
Expanding both sides: \(x^2 - 6x + 9 = x^2 - 5x\).
Simplifying: \(-6x + 9 = -5x\).
Rearranging gives: \(x = 9\).
(ii) The common ratio \(r\) is \(\frac{x-3}{x} = \frac{6}{9} = \frac{2}{3}\).
The fourth term is \(ar^3 = 9 \times \left(\frac{2}{3}\right)^3 = 9 \times \frac{8}{27} = \frac{72}{27} = 2\frac{2}{3}\) or 2.67.
(iii) The sum to infinity \(S_\infty\) is given by \(\frac{a}{1-r}\).
Substituting the values: \(S_\infty = \frac{9}{1 - \frac{2}{3}} = \frac{9}{\frac{1}{3}} = 27\).
