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A geometric sequence has $400$ terms. The first term is $1000$ and the common ratio is $-\frac{8}{9}.$ How many terms of this sequence are greater than $1?$

 May 9, 2024
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\(1000 \left(-\dfrac89\right)^{n-1} > 1\)

To be greater than 1, it must be greater than 0, so we only count the odd numbered terms, i.e. 1st term, 3rd term, etc.

After some point, even if it is an odd numbered term, it won't be greater than 1 anymore.

Since the term is an odd numbered term as we deduced, we can safely let n = 2k + 1 for k >= 0, k is an integer.

 

\(1000 \left(\dfrac{64}{81}\right)^k > 1\\ k < \log_{64/81}{\dfrac1{1000}}\\ k < 29.3\)

 

Then any nonnegative integer k with k < 29 will correspond to a positive term in the original sequence which is greater than 1.

 

Hence, the answer is 29.

 May 10, 2024

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