Question

the sum of three consecutive terms of a gp is 13/2 and their product is -1.find the terms​

Answer 1

Answer:

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