# If the sum of three numbers in G.P is 38, and their product is 1728, find the numbers.

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sum is ⇒ a+ar+ar² = 38 ----------- (i)

product is⇒ a*ar*ar² = 1728 ---------- (ii)

⇒ a³r³ = 1728

⇒ (ar)³ = (12)³

⇒

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,

we get

hence, the numbers are

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

OR