What are the three nunbers in a geometric progression whose sum is 10.1/2 and product is 27?

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Answer 1 · 2018-02-05T03:42:05

# (1): 3/2,3,6#.

Suppose that, the reqd. nos. ( in a GP ) are, #a/r,a,ar#.

Then, by what is given, we have,

#a/r+a+ar=10 1/2=21/2...(1), and a/r*a*ar=27...(2)#.

#(2)rArr a^3-27=0 rArr (a-3)(a^2+3a+9)=0#.

#:. a=3, or, a=[-3+-sqrt{(-3)^2-4*1*9}]/(2*1); i.e., #

# a=3, or, a=3((-1+-isqrt3)/2), i.e., #

# a=3, a=3omega, or a=3omega^2; ...[omega"=cube root of unity]"#.

Case 1: #a=3#.

From #(1)#, we get, #3(1/r+1+r)=21/2, or, 2(1+r+r^2)=7r, i.e.,#

#2r^2-5r+2=0 rArr r=2, or r=1/2#.

#:. a=3 and r=2" give the desired nos. : "3/2,3,6;" while "#

#a=3, r=1/2" give : "6,3,3/2#.

The Remaining Cases can be dealt with similarly.