Geometric progression
posted by Anonymous .
The second term of a geometric progression is 12 more than the first term given that the common the ratio is half of the first term. Find the third term of the Geometric progression

Solution.Tn=ar'n1 where n=3

ar = 12 + a (i)
ar^2 = x (ii)
subtract equ (i) from equ (ii)
a = 6
substituting a = 6 into equ (i)
ar = 12 + a
6r = 12 + 6=18
r = 3.
ar'2 = 6 * 3'2
ar'2 = 54 
65

ar'2=54