The first term in a progression is 36 and the second term is 32.
(i) Given that the progression is geometric, find the sum to infinity.
(ii) Given instead that the progression is arithmetic, find the number of terms in the progression if the sum of all the terms is 0.
Solution
(i) For a geometric progression, the common ratio is given by \(r = \frac{32}{36} = \frac{8}{9}\).
The sum to infinity for a geometric series is \(S_\infty = \frac{a}{1-r}\), where \(a = 36\) and \(r = \frac{8}{9}\).
Thus, \(S_\infty = \frac{36}{1 - \frac{8}{9}} = \frac{36}{\frac{1}{9}} = 36 \times 9 = 324\).
(ii) For an arithmetic progression, the common difference is \(d = 32 - 36 = -4\).
The sum of an arithmetic series is given by \(S_n = \frac{n}{2} (2a + (n-1)d)\).
Given \(S_n = 0\), we have:
\(0 = \frac{n}{2} (72 + (n-1)(-4))\)
\(0 = \frac{n}{2} (72 - 4n + 4)\)
\(0 = \frac{n}{2} (76 - 4n)\)
\(0 = n(38 - 2n)\)
\(0 = 38n - 2n^2\)
\(2n^2 = 38n\)
\(n(2n - 38) = 0\)
\(n = 0\) or \(n = 19\)
Since \(n = 0\) is not a valid solution, \(n = 19\).
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