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The sum of infinity of a convergent series is 243. The sum of the first five terms is 242. Determine the common ration and the first term
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\(From, \space S_\infty = \frac{a}{1-r} \space we \space have,\\ 243 = \frac{a}{1-r}\\ \implies a = 243(1-r)\\ \implies a = -243(r-1) \space \space \space ... \space(1)\\ \\ and \space from \space S_n = \frac {a(r^n-1)}{r-1}, \space we \space have\\ 242 = \frac{a(r^5-1)}{r-1} \space \space \space ... \space(2)\\ \\ subt. (1) \space into \space (2), we \space have\\ 242 = \frac{-243(r-1)(r^5-1)}{r-1} \\ \implies 242 = -243(r^5-1)\\ -\frac{242}{243} = r^5 -1 \\ r^5 =\frac{1}{243}\\ r = \frac{1}{3}\\ subt. r \space into\space (1) \implies a = 243(1-\frac{1}{3})\\ a = 162\)

by Wooden (882 points)

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