Second term of a GP is 1000 and the common ratio is r = (1/n) 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. 4
B. 9
C. 5
D. 13
Answer: 9
Common ratio is positive, and one of the terms is positive => All terms are positive
P6 = P5 * t6 => If P6 > P5, t6 > 1
P7 = P6 * t7 => If P6 > P7, t7 < 1
t6 = t2 * r^4 = 1000r^4;
t7 = t2 * 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