Question

31. Find the three numbers in GP whose sum is 39 and their product is 729.​

Answer 1

Answer:

39 = ar + ar^2 + ar^3  

729 = (ar) * (ar^2) * (ar^3)  

729 = (ar) * (ar^2) * (ar^3)  

729 = (ar)^3 * r^3  

9^3 = (ar)^3 * r^3  

9 = ar * r  

9 = ar^2  

39 = ar + ar^2 + ar^3  

39 = ar^2 / r + ar^2 + ar^2 * r  

39 = 9/r + 9 + 9r  

13 = 3/r + 3 + 3r  

13r = 3 + 3r + 3r^2  

0 = 3 - 10r + 3r^2  

r = (10 +/- sqrt(100 - 36)) / 6  

r = (10 +/- 8) / 6  

r = 18/6 , 2/6  

r = 3 , 1/3  

ar^2 = 9  

a * 3^2 , a * (1/3)^2 = 9  

a * 9 , a * (1/9) = 9  

a = 1 , 81  

If r = 3, then:  

ar , ar^2 , ar^3 =>  

1 * 3 , 1 * 3^2 , 1 * 3^3 =>  

3 , 9 , 27  

If r = 1/3, then:  

ar , ar^2 + ar^3 =>  

81 / 3 , 81 / 9 , 81 / 27 =>  

27 , 9 , 3  

Either way, the 3 numbers are 3 , 9 , and 27.

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