Question 1070586
The nth term of an arithmetic series can be written a_n = a + (n-1)d where 
a is the 1st term and d is the common difference
Given that the 1st term minus the 3rd term is 8, we have
a - (a + 2d) = 8 -> -2d = 8, or d = -4
So a_n = a - 4(n-1)
Adding 3, 5 and 8 to the 1st three terms gives g_1 = a + 3, g_2 = a + 1, and g_3 = a
Since the ratio of successive terms of a geometric sequence is a constant, r,  we can write:
r = (a+1)/(a+3) = a/(a+1)
Solve for a: 
a^2 + 2a + 1 = a^2 + 3a -> a = 1
The arithmetic sequence is a_n = 1 - 4(n-1)
The common ratio is 1/(1+1) = 1/2
So the 1st 4 terms of the geometric sequence are 4, 2, 1, 1/2