SOLUTION: Find three number of G.P whose sum is 19 and product is 216

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Question 1028877: Find three number of G.P whose sum is 19 and product is 216
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
Let the numbers be a/r, a,ar

a/r + a +r =19
a/r*a*ar=219
a^3= 216
a=6
a+ar+ar^2 = 19r

6+6r+6r^2=19r
6r^2-13r+6 =0
6r^2-9r-4r+6=0
3r(2r-3)-2(2r-3)=0
(2r-3)(3r-2)=0
r=3/2 OR r=2/3
plug a & r
when r=3/2
a/r = 6/(3/2) =4
a=6
ar = 9
4,6,9
OR r=2/3
a/r = 6/(2/3) =9
a= 6
ar = 4
9,6,4

Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.
Find three number of G.P whose sum is 19 and product is 216
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Answer.   1) a = 9, r = , and the progression is {9, 6, 4};   2) a = 4, r = , and the progression is {4, 6, 9}.

Solution 

Let "a" be the first term of the GP,  and "r" be the common ratio.
Then the three terms of the GP are ,  and . 
From the condition, we have these two equations

 = ,    (1)

 = .   (2)

From (2), you have

 = ,  hence,   =  = 6.   (3)


   (I consider here only real solutions and do not consider complex values).


Based on (3), substitute 6 into (1) instead of the second addend. You will get

 = ,   or

 =  = .   (4)

Now, from (3) express r =  and substitute it into (4). You will get

 = ,  or

 = .    (5)

Multiply both sides of (5) by a to red off the denominator. You will get

 = ,   or    = .   (6)

Factor the left side of the equation (6). You will get

(a-9)*(a-4) = 0

with the roots a = 9  or  a = 4.
Then the corresponding values of  "r"  are  r =  =   and  r =  = ,  exactly as in the answer above.

Thus the progression is  {9, 6, 4}  or  {4, 6, 9}.


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