For what values of t will t-2, 2t-6, and 4t-8, in this order, form an arithmetic sequence?

Question

Could someone at least start me off?

Answer 1

There is an arithmetic progression if
2t-6 -(t-2) = (4t-8)-(2t-6)
t -4 = 2t -2
t = -2
in which case the sequence is -4, -10, -16

Answer 2

Well calculated

Answer 3

t-2,2t-6,4t-8 =(4t-8)=(2t-6)=4t-2t=-8+6=2t=-2=(t=-1)is the same:2t-6=t-2(t=-3)(-3= t=-1)(t=-3+1)t=-2 where is -4,-10,-16