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 , 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


  • Correct Answer
    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

< 1000 and > 1000.
= 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

  1. Counting and Progressions
  2. Common Ratio
  3. Common Difference
  4. Sum up to 'n' Terms
  5. AP Puzzle
  6. GP Logical Puzzle
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.