Question

Find three numbers in G.P. such that their product is 216 and the sum of first and third numbers is 20.

Answer 1

Solution :

Let a/r , a , ar are three terms in G.P.

i ) Product = 216 ( given

=> ( a/r ) × a × ar = 216

=> a³ = 6³

=> a = 6 ----( 1 )

ii ) Sum of first term and third term = 20

=> a/r + ar = 20

=> 6/r + 6r = 20 [ from ( 1 ) ]

Divide each term by r/2 , we get

=> 3 + 3r² = 10r

=> 3r² - 10r + 3 = 0

Splitting the middle term, we get

=> 3r² -9r - 1r + 3 = 0

=> 3r( r - 3 ) - 1( r - 3 ) = 0

=> ( r - 3 )( 3r - 1 ) = 0

=> r - 3 = 0 or 3r - 1 = 0

=> r = 3 or r = 1/3

Therefore ,

i ) If a = 6 , r = 3

Required G.P is 6/3 , 6 , 6×3

Or

2 , 6 , 18

ii ) If a = 6 , r = 1/3

Required G.P ,

18 , 6 , 2

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