3. For a linear sequence, T(5) = 18 and T(7) = 12 a) Find T(1) b) Find T(100)

First you know that when x=5 T= 18 and x=7 T=12
then you can use the gradient formula which is
g= (y2-y1)/(x2-x1)

let x1=5 ,y1 = 18, x2=7 y2=12

therefore,
g= (12-18)/(7-5)
g= -6/2
g= -3

hence 
y= -3x + c
when x = 5 y = 18
18 = -15 + c
c= 33

T= -3x+33

when x = 1
T = -3 + 33
T = 30

when x = 100
T= -300 + 33
T =  -267