Answers

2016-04-25T11:03:00+05:30
Let the numbers be  a, ar, ar²
 sum is ⇒  a+ar+ar² = 38     -----------  (i)
product is⇒ a*ar*ar² = 1728   ---------- (ii)
               ⇒  a³r³ = 1728 
               ⇒  (ar)³ = (12)³
             ⇒    ar = 12    a = 12/r  --------- (iii)
frm (i) 
  12/r + 12 + 12r²/r = 38 
   12r² - 26r + 12 = 0
 taking 2 common
 6r² - 13r + 6 = 0
 6r² -4r - 9r + 6 =0
 ∴ r = 2/3,  r = 3/2
putting, r= 3/2  in (iii)
we get  a = 8
hence, the numbers are
8,12,18 
1 5 1
2016-04-25T11:05:49+05:30
Let the numbers be a/r , a and ar.
According to the question ,
a/r × a × ar = 1728.
a³ = 1728
a³ = (12)³³
a = 12   [1]

a/r + a + ar = 38
a[ 1/r +1+ r ] = 38
12[ 1 + r + r² ] / r = 38 { using 1 }

12 + 12r + 12r² = 38r
12 + 12r² = 38r - 12r = 26r
12 - 26r + 12r² = 0
12r² - 26r +12 = 0
12r² - 18r - 8r + 12 = 0
6r( 2r - 3 ) - 4( 2r - 3 ) = 0
( 6r - 4 )( 2r - 3 ) = 0

r = 2/3 , 3/2

the numbers are 

18 , 12 , 8

      OR

8 , 12 , 18
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