# Arithmetic and Geometric Progressions

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This question is almost a puzzle. You have to carefully consider the implications for the common ratio, and then extrapolate to the Product of n terms.

## Geometric Progressions

Second term of a GP is 1000 and the common ratio is $r = \frac{1}{n}$ , 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?
1. 4
2. 9
3. 5
4. 13

Choice B. 9

## Detailed Solution

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 * r4 = 1000r4;

t7 = t2 * r5 = 1000r5

1000r4 > 1 and 1000r5 < 1

${\frac{1}{ r^{4} }}$ < 1000 and ${\frac{1}{ r^{5} }}$ > 1000.
${\frac{1}{r}}$ = n.

n4 < 1000 and n5 > 1000, where n is a natural number

n4 < 1000 => n < 6

n5 > 1000 => n ≥ 4

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

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## More questions from Progressions

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. Reinforce these ideas with these questions.