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In the sequence \(a_1, a_2, \ldots, a_n,\ldots,\) we have \(a_1=x\) and \(a_n=y-z\cdot a_{n-1}\) for all \(n>1.\) Is

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In the sequence \(a_1, a_2, \ldots, a_n,\ldots,\) we have \(a_1=x\) and \(a_n=y-z\cdot a_{n-1}\) for all \(n>1.\) Is

by VJesus12 » Thu Nov 04, 2021 6:26 am

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In the sequence \(a_1, a_2, \ldots, a_n,\ldots,\) we have \(a_1=x\) and \(a_n=y-z\cdot a_{n-1}\) for all \(n>1.\) Is \(a_3>a_2?\)

(1) \(z > y^2 + 2\)
(2) \(x > \dfrac{y}{z+1}\)

Answer: C

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