the sum of three consecutive terms of a gp is 13/2 and their product is -1.find the terms
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a(1+r+r²)=38
(ar)³=1728=12³
ar=12
a+ar+ar²=38
a+12+12r=38
a+12r=26
12/r+12r=26
6/r+6r=13
6r²—13r+6=0
(3r—2)(2r—3)=0
r=2/3 or 3/2
a=12*3/2=18, or 12*2/3=12
GP is
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