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New post 08 Oct 2017, 02:58
A sequence of Q consists of 15 numbers arranged in ascending order.the first term in this sequence is 25. In this sequence for the first 14 terms the ratio of the term to the next term is a fixed constant. Last term of the sequence is four times the first term. What is the 8th term in the sequence ?

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New post 08 Oct 2017, 03:16
Please share options. This question would probably need eliminating wrong choices.

The sequence Q could be:\(25, 25a, 25a^2, 25a^3, .... 25a^(13), 100\)
where a>1

Since we know that all the terms are in ascending order, 14th term would be less than 15th term.
\(25a^(13) < 100\)
\(a^13 < 4\)
\(a < 2^\frac{2}{13}\)


8th term would be \(25a^7\)
8th term < \(25 * 2^\frac{14}{13}\)
8th term < \(25 * 2 * 2^\frac{1}{13}\)
8th term < \(50 * 2^\frac{1}{13}\)

Now look at the options to find a close match 25 - 53
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New post 08 Oct 2017, 03:20
And is 50 abhishek..

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New post 08 Oct 2017, 03:21
Sorry, no options available..

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New post 08 Oct 2017, 03:22
Correct ans is 50

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New post 21 Oct 2017, 09:02
kunal4799 wrote:
A sequence of Q consists of 15 numbers arranged in ascending order.the first term in this sequence is 25. In this sequence for the first 14 terms the ratio of the term to the next term is a fixed constant. Last term of the sequence is four times the first term. What is the 8th term in the sequence ?

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Re: Progressions.   [#permalink] 21 Oct 2017, 09:02
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