Arithmetic and geometric progression

  • Thread starter tykescar
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  • #1
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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?
 

Answers and Replies

  • #2
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.
 
  • #3
HallsofIvy
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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.
 
  • #4
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Got it, thanks.
 

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