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Let \(a_1, a_2, \dots, a_{99}\) be a geometric sequence.  If \(a_{49} = 18\) and \(a_{51} = 8\) then find the geometric mean of all 99 terms.

 Jul 6, 2020
 #1
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so the first sequence is (18-8)*48+18=498 and the last number is -232 so the formula is (498+-232)*99/2=13167 so the sum of the 99 terms is 13167

 Jul 6, 2020
 #2
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This is a geometric sequence with a first term =717897987691852588770249/140737488355328 
Common Ratio = 2/3
Number of terms =99
Sum = [717897987691852588770249/140737488355328] * (1 - (2/3)^99) / (1 - 2/3), solve for S
S =1 5302916005.13719
The geometric mean =S^(1/99) =1.2672914563....

 Jul 6, 2020
 #3
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Guest #2: You started OK, but I think you went astray towards the end. The geometric mean is 1/99th root of the products of all the terms of the sequence (when multiplied together). So, based on that, this is what I got as the "geometric mean" of the 99 terms of this sequence:

 

productfor(n, 0, 98, (717897987691852588770249/140737488355328)*(2/3)^n=(6.901497877 E+106)^(1/99)=12  - which is the geometric mean of the 99 terms of this sequence.

 Jul 6, 2020

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