Arithmetic and geometric progression

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Discussion Overview

The discussion revolves around a problem involving arithmetic progression (AP) and geometric progression (GP). Participants explore the conditions under which specific terms of an AP can be related to terms of a GP, and how to derive the common difference of the AP and the common ratio of the GP based on given constraints.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant initially misunderstands the problem, assuming the fourth, seventh, and sixteenth terms of the AP are also the terms of the GP.
  • Another participant clarifies that the problem states the fourth, seventh, and sixteenth terms of the AP correspond to the first, second, and third terms of the GP, respectively.
  • A third participant formulates the terms of the AP in terms of its first term (a) and common difference (d), establishing equations based on the properties of the GP and the sum of the first six terms of the AP.
  • The equations derived include relationships between the terms of the AP and the GP, leading to a system of equations to solve for a, d, and the common ratio (r).

Areas of Agreement / Disagreement

Participants generally agree on the interpretation of the problem after clarification, but the discussion remains focused on deriving the solution without reaching a consensus on the values of a, d, and r.

Contextual Notes

The discussion involves assumptions about the relationships between the terms of the AP and GP, and the need to solve a system of equations that has not yet been fully resolved.

tykescar
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If the fourth, seventh and sixteenth terms of an AP are in geometric progression, the first six terms of the AP have a sum of 12, find the common difference of the AP and the common ratio of the GP.
I've been assuming that the fourth, seventh and sixteenth terms of the AP are the fourth, seventh and sixteenth of the GP, which isn't the case.
How do I start?
 
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Hi,
actually the question says that the 4th, 7th and 16th term of the AP is the 1st, 2nd and 3rd term of the GP.
Then it is solvable.
 
Let a be the first term, d the common difference. Then the 4th, 7th, and 16th terms are a+ 3d, a+ 6d, and a+ 15d. Saying that they form a geometric progression means that a+ 6d= r(a+ 3d) and a+ 15d= r(a+ 6d). Also, the 6th term is a+ 5d so the sum of the first 6 terms is ((a+ a+ 5d)/2)(6)= 3(2a+ 5d)= 12. That gives three equations to solve for a, d, and r.
 
Got it, thanks.
 

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