Asked by Archiebald
                three consecutive terms of a geomentric progression series have product 343 and sum 49/2.  
fine the numbers.
HOW WILL ONE SOLVE THAT?
THANKS
            
        fine the numbers.
HOW WILL ONE SOLVE THAT?
THANKS
Answers
                    Answered by
            Reiny
            
    just use your definitions:
the terms would be a, ar, and ar^2
so a(ar)(ar^2) = 343
a^3 r^3 = 343
(ar)^3 = 343
ar = 7 or a = 7/r
a + ar + ar^2 = 49/2
a(1 + r + r^2) = 49/2
(7/r)(1 + r + r^2) = 49/2
7/r + 7 + 7r = 49/2
times 2r
14 + 14r + 14r^2 = 49r
14r^2 - 35r + 14 = 0
2r^2 - 5r + 2 = 0
(2r - 1)(r - 2) = 0
r = 1/2 or r = 2
if r = 1/2, a = 7/(1/2) = 14
if r = 2, a = 7/2
state your conclusion
    
the terms would be a, ar, and ar^2
so a(ar)(ar^2) = 343
a^3 r^3 = 343
(ar)^3 = 343
ar = 7 or a = 7/r
a + ar + ar^2 = 49/2
a(1 + r + r^2) = 49/2
(7/r)(1 + r + r^2) = 49/2
7/r + 7 + 7r = 49/2
times 2r
14 + 14r + 14r^2 = 49r
14r^2 - 35r + 14 = 0
2r^2 - 5r + 2 = 0
(2r - 1)(r - 2) = 0
r = 1/2 or r = 2
if r = 1/2, a = 7/(1/2) = 14
if r = 2, a = 7/2
state your conclusion
                    Answered by
            Obed
            
    How did the 2r come about?
    
                    Answered by
            Thomas
            
    Is that all
    
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.