Question

the sum of first 3 terms of a gp is 13/12 and their product is - 1 find gp

Answer 1

let the first 3 terms be a/r+a+ar
""we have taken these terms because when we'll multiply them "r" will get cancelled.""

s3=13/12
a/r+a+ar=13/12.....(1)
a/r×a×ar=(-1)
(r WILL GET CANCELLED)
a^3 =(-1)
a=(-1)....(2)

NOW PUT THE VALUE OF a IN EQUATION (1)

(-1)/r+(-1)+(-1)r=13/12
-1/r-r=13/12+1
now taking lcm

-1-r^2/r =13+12/12

-1-r^2/r=25/12

now cross multiplication

(-1-r^2)×12=25×r
-12-12r^2=25r
12r^2+25r+12=0
middle splitting the terms
(12r^2+9r)+(16r+12)=0
3r(4r+3)+4(4r+3)=0
(4r+3) (3r+4)

now taking (4r+3)=0
4r+3=0
4r=(-3)
r= -3/4

now taking (3r+4)=0
3r+4=0
3r= -4
r= -4/3

i hope this will help