### CAT Progressions

Question

Second
term of a GP is 1000 and the common ratio is
where n is a natural number. Pn is the product of n terms of this GP. P6
> P5 and P6 > P7, what is the sum of all possible values of n?

A. 5

B. 9

C. 8

D. 11

Correct Answer

Choice B

Explanatory Answer

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
1/r^4 < 1000 and 1/r^5 > 1000

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

**Answer Choice (B)**

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