The question is about CAT Geometric Progressions. A question on possible values. We need to find out the possible values of 'n' which is related to the common ratio of a GP. With some simple but very powerful ideas, one can cut down on a lot of working when it comes to progressions. For example, anchoring a progression around its middle term can be very useful. CAT Exam tests the idea of progression often in the CAT Quantitative Aptitude section and this could also be tested in DI LR section of the CAT Exam as a part of a puzzle.

Question 1: Second term of a GP is 1000 and the common ratio is r = \\frac{1}{n}\\) where n is a natural number. P_{n} is the product of n terms of this GP. P_{6} > P_{5} and P_{6} > P_{7}, what is the sum of all possible values of n?

- 4
- 9
- 5
- 13

9

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Common ratio is positive, and one of the terms is positive => All terms are positive

P_{6} = P_{5} * t_{6} => If P_{6} > P_{5}, t_{6} > 1

P_{7} = P_{6} * t_{7} => If P_{6} > P_{7}, t_{7} < 1

t_{6} = t_{2} * r^{4} = 1000r^{4};

t_{7} = t_{2} * r^{5} = 1000r^{5}

1000r^{4} > 1 and 1000r^{5} < 1

\\frac{1}{r^{4}}\\) < 1000 and \\frac{1}{r^{5}}\\) > 1000.

\\frac{1}{r}\\) = n.

n^{4} < 1000 and n^{5} > 1000, where n is a natural number

n^{4} < 1000 => n < 6

n^{5} > 1000 => n ≥ 4

n could be 4 or 5. Sum of possible values = 9

The question is **"what is the sum of all possible values of n?"**

Choice B is the correct answer.

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