The recursion relation for a geometric sequence is an = an-1*r. The sum of the first two terms is:
a0 + a1 =a0 + a0*r = a0(1+r)
The sum of the first 4 terms is:
a0 + a1 + a2 + a3 = a0 + a0*r + a0*r2 + a0*r3 = a0(1+r+r2+r3)
The sum of the first 4 terms equals 5 times the sum of the first 2 terms:
a0(1+r+r2+r3) = 5a0(1+r)
1 + r + r2 + r3 = 5 + 5r
r3 + r2 - 4r - 4 = 0
Factor by grouping:
r2(r+1) - 4(r+1) == 0
(r2-4)(r+1) = 0
(r+2)(r-2)(r+1) = 0
r = -2, 2, and -1
Since the problem states that this is a positive sequence we can eliminate the negative values for r, leaving r = 2.
Check:
1 + r + r2 + r3 = 5(1+r)
(1 + 2 + 22 + 23) = 5(1 + 2)
(1 + 2 + 4 + 8) = 5(3)
15 = 15
Check!
