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If k+1,2k-1 and 3k+1 are three consecutive numbers of a geometric progression what is...

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christiano-cr7 | Salutatorian

Posted June 27, 2013 at 5:36 AM via web

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If k+1,2k-1 and 3k+1 are three consecutive numbers of a geometric progression what is the possible answers for common ratio.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted June 27, 2013 at 5:45 AM (Answer #1)

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By the three numbers we can derive the common ratio(r) as;

`r = (2k-1)/(k+1) = (3k+1)/(2k-1)`

 

`(2k-1)/(k+1) = (3k+1)/(2k-1)`

`(2k-1)(2k-1) = (3k+1)(k+1)`

`4k^2-4k+1 = 3k^2+4k+1`

`k^2-8k = 0`

`k(k-8) = 0`

 

When k = 0 then r `=(2xx0-1)/(0+1) = -1`

When k = 8 then r `= (2xx8-1)/(8+1) = 15/9 = 5/3`

 

So the possible values for common ratio is -1 or 5/3.

 

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