Question 1112832
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If the common ratio is *[tex \Large r], and the *[tex \Large n]th term is *[tex \Large t_n] and the *[tex \Large m]th term is *[tex \Large t_m], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{t_m}{t_n}\ =\ r^{m-n}]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{25}{400}\ =\ r^{5-3}\ =\ r^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2\ =\ \frac{1}{16}\ \Right\ r\ =\ \frac{1}{4}]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{t_8}{t_5}\ =\ \(\frac{1}{4}\)^3\ =\ \frac{1}{64}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{t_8}{25}\ =\ \frac{1}{64}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t_8\ =\ \frac{25}{64}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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