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Nov 2009 p11 q8
808
The first term of an arithmetic progression is 8 and the common difference is \(d\), where \(d \neq 0\). The first term, the fifth term and the eighth term of this arithmetic progression are the first term, the second term and the third term, respectively, of a geometric progression whose common ratio is \(r\).
(i) Write down two equations connecting \(d\) and \(r\). Hence show that \(r = \frac{3}{4}\) and find the value of \(d\). [6]
(ii) Find the sum to infinity of the geometric progression. [2]
(iii) Find the sum of the first 8 terms of the arithmetic progression. [2]
Solution
(i) The first term of the arithmetic progression is 8, so the first term of the geometric progression is also 8. The fifth term of the arithmetic progression is \(8 + 4d\), which is the second term of the geometric progression, so \(8 + 4d = 8r\). The eighth term of the arithmetic progression is \(8 + 7d\), which is the third term of the geometric progression, so \(8 + 7d = 8r^2\).
We have the equations:
\(8 + 4d = 8r\)
\(8 + 7d = 8r^2\)
Eliminating one of the variables, we solve:
\(4r^2 - 7r + 3 = 0\)
Solving this quadratic equation gives \(r = \frac{3}{4}\) and \(d = -\frac{1}{2}\).
(ii) The sum to infinity of the geometric progression is given by \(S_\infty = \frac{a}{1-r}\), where \(a = 8\) and \(r = \frac{3}{4}\).
\(S_\infty = \frac{8}{1 - \frac{3}{4}} = 32\)
(iii) The sum of the first 8 terms of the arithmetic progression is given by \(S_8 = \frac{n}{2} (2a + (n-1)d)\), where \(n = 8\), \(a = 8\), and \(d = -\frac{1}{2}\).
