Answer:
The inverse of
is 
Step-by-step explanation:
Given equation :

Replace x with y and y with x
So, 
So, Option C is correct
Hence The inverse of
is 
Do u have a pic of the problem ?
Step-by-step explanation:
a1=ar⁰=12 A2=ar¹=-6
a=12 ar=-6
12r=-6
r=-6/12
r=-1/2
a10=ar⁹
=12*(-1/2)⁹
=12*(-512)
=-3/128
S∞ = a1 / (1-r )
=12/(1-1/2)
=12*1/2
=6