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The second term of a GP is 1000 and the common ratio is r = 1/n, where

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The second term of a GP is 1000 and the common ratio is r = 1/n, where  [#permalink]

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New post 11 Nov 2019, 23:38
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The second term of a GP is 1000 and the common ratio is r = 1/n, where n is a natural number. P(n) is the product of n terms of this GP series. If P(6)>P(5) and P(6)>P(7), what is the sum of all possible values of n?

A. 4
B. 5
C. 9
D. 13
E. 15
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Re: The second term of a GP is 1000 and the common ratio is r = 1/n, where  [#permalink]

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New post 11 Nov 2019, 23:59
If the second term is 1000, then the first term is 1000n
T6 = 1000n/n^5 = 1000/n^4
T7 = 1000n/n^6 = 1000/n^5

Now, P(6) > P(5)
So, T1*T2*T3*T4*T5*T6 > T1*T2*T3*T4*T5
T6>1
1000/n^4 > 1
So, n^4 < 1000

Similarly, P(6)>P(7)
So, T1*T2*T3*T4*T5*T6 > T1*T2*T3*T4*T5*T6*T7
Hence,
T7 < 1
1000/n^5 < 1
n^5> 1000

Finally,
n^4 < 1000 < n^5

Only, 4 and 5 satisfy this range

Hence, sum of values of N is 9

Option C

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Re: The second term of a GP is 1000 and the common ratio is r = 1/n, where   [#permalink] 11 Nov 2019, 23:59
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The second term of a GP is 1000 and the common ratio is r = 1/n, where

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