Asked by blake
                if (3-x)=6, (7-5x) are consecutive terms of GP with common ratio r>0, fing d value of (x) and d common ratio. 
thanks
            
        thanks
Answers
                    Answered by
            blake
            
    R = u2/u1 = u2/u2 = u4/u3 
6/3-x = 7-5x/6
    
6/3-x = 7-5x/6
                    Answered by
            Reiny
            
    You will need brackets in
6/(3-x) = (7-5x)/6
and
(3-x)(7-5x) = 36
21 -22x + 5x^2 = 36
5x^2 - 22x - 15 = 0
(x - 5)(5x + 3) = 0
x = 5 or x = -3/5
the common ratio is 6/(3-x)
if x =5, common ratio is 6/-2 or -3
if x = -3/5, common ratio is 6/(3 + 3/5) or 5/3
    
6/(3-x) = (7-5x)/6
and
(3-x)(7-5x) = 36
21 -22x + 5x^2 = 36
5x^2 - 22x - 15 = 0
(x - 5)(5x + 3) = 0
x = 5 or x = -3/5
the common ratio is 6/(3-x)
if x =5, common ratio is 6/-2 or -3
if x = -3/5, common ratio is 6/(3 + 3/5) or 5/3
                    Answered by
            Mary
            
    Fayokemi
    
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