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June 2012 p13 q6
803
The first term of an arithmetic progression is 12 and the sum of the first 9 terms is 135.
(i) Find the common difference of the progression.
The first term, the ninth term and the nth term of this arithmetic progression are the first term, the second term and the third term respectively of a geometric progression.
(ii) Find the common ratio of the geometric progression and the value of n.
Solution
(i) The sum of the first \(n\) terms of an arithmetic progression is given by \(S_n = \frac{n}{2} (2a + (n-1)d)\).
For the first 9 terms, \(S_9 = \frac{9}{2} (24 + 8d) = 135\).
Solving for \(d\), we have:
\(9(24 + 8d) = 270\)
\(24 + 8d = 30\)
\(8d = 6\)
\(d = \frac{3}{4}\).
(ii) The 9th term of the arithmetic progression is \(12 + 8 \times \frac{3}{4} = 18\).
In the geometric progression, the first term is 12, the second term is 18.
The common ratio \(r\) is \(\frac{18}{12} = \frac{3}{2}\).
The third term of the geometric progression is \(ar^2 = 12 \times \left(\frac{3}{2}\right)^2 = 27\).
The \(n\)th term of the arithmetic progression is \(12 + (n-1) \times \frac{3}{4} = 27\).
