Question 974769
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I presume you mean


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a_n\ =\ \frac{a_{n-1}}{a_{n-2}}]


If you work out the first 12 terms, you will notice a pattern:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2,\ 3,\ \frac{3}{2},\ \frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ 2,\ 3,\ \frac{3}{2},\ \frac{1}{2},\ \frac{1}{3},\ \frac{2}{3}]


Yes, indeed, the set of values repeats every six terms.  Using integer division, divide 1485 by 6 and note the remainder.  Count the number of terms in the list up to the remainder you just obtained, and that will be the value of *[tex \LARGE a_{1485}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \