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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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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.
Posted by jeew-m on June 27, 2013 at 5:45 AM (Answer #1)
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