The first, second and third terms of a geometric progression are \(2p + 6\), \(5p\) and \(8p + 2\) respectively.
(a) Find the possible values of the constant \(p\).
(b) One of the values of \(p\) found in (a) is a negative fraction. Use this value of \(p\) to find the sum to infinity of this progression.
Solution
(a) For a geometric progression, the ratio between consecutive terms is constant. Therefore, we have:
\(\frac{5p}{2p + 6} = \frac{8p + 2}{5p}\)
Cross-multiplying gives:
\(5p(5p) = (2p + 6)(8p + 2)\)
\(25p^2 = 16p^2 + 12p + 48p + 12\)
\(25p^2 = 16p^2 + 60p + 12\)
Rearranging gives:
\(9p^2 - 60p - 12 = 0\)
Factoring gives:
\((9p + 2)(p - 6) = 0\)
Thus, \(p = \frac{-2}{9}\) or \(p = 6\).
(b) Using \(p = \frac{-2}{9}\), the first term \(a\) is:
\(a = 2\left(\frac{-2}{9}\right) + 6 = \frac{50}{9}\)
The common ratio \(r\) is:
\(r = \frac{5p}{2p + 6} = \frac{10/9}{50/9} = \frac{-1}{5}\)
The sum to infinity \(S_\infty\) is given by:
\(S_\infty = \frac{a}{1 - r} = \frac{50/9}{1 - (-1/5)} = \frac{50/9}{6/5} = \frac{125}{27}\)
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