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Nov 2016 p13 q9
904
Two convergent geometric progressions, P and Q, have the same sum to infinity. The first and second terms of P are 6 and 6r respectively. The first and second terms of Q are 12 and -12r respectively. Find the value of the common sum to infinity.
Solution
The sum to infinity of a geometric progression is given by \(S = \frac{a}{1 - r}\), where \(a\) is the first term and \(r\) is the common ratio.
For progression P, the first term \(a = 6\) and the second term is \(6r\), so the common ratio \(r\) is \(r\).
The sum to infinity for P is \(S_P = \frac{6}{1 - r}\).
For progression Q, the first term \(a = 12\) and the second term is \(-12r\), so the common ratio \(r = -r\).
The sum to infinity for Q is \(S_Q = \frac{12}{1 + r}\).
Since the sums are equal, \(\frac{6}{1 - r} = \frac{12}{1 + r}\).
Cross-multiplying gives:
\(6(1 + r) = 12(1 - r)\)
\(6 + 6r = 12 - 12r\)
\(18r = 6\)
\(r = \frac{1}{3}\)
Substitute \(r = \frac{1}{3}\) back into the sum formula for P:
