Question 1083343
The nth term of the arithmetic progression is a_n = a + (n-1)d, where a is the 1st term and d is the common difference
The nth term of the geometric progression is a_n = ar^(n-1), where a is the 1st term and r is the common ratio.
Second term of the G.P. = third term of the A.P. -> ar   = a + 2d   [1]
Third term of the G.P. = fourth term of the A.P. -> ar^2 = a + 3d   [2]
Eliminate d by multiplying [1] by 3/2 and subtracting [1] from [2]:
ar^2 - (3/2)ar = a - (3/2)a
Dividing through by a and simplifying gives:
r^2 - (3/2)r + 1/2 = 0 -> 2r^2 - 3r + 1 = 0
This gives two solutions, r = 1 and r = 1/2
r = 1 makes all terms equal to the first term, so we choose the 2nd solution
Ans: r = 1/2
Using [1] and the value for r, we have:
(1/2)a = a + 2d -> a = -4d
The 5th term of the A.P. is a_5 = a + 4d -> a_5 = -4d + 4d = 0