Problem: The first three terms of a geometric progression are 2β,32β,62β. The fourth term is:
Answer Choices:
A. 1
B. 72β
C. 82β
D. 92β
E. 102β
Solution:
1=arnβ1,r=32βΓ·2ββ΄ the 4th term is 2β(2β32ββ)3=22β2ββ
2β=1
or
Since 2β,32β,62β can be written as 23/6,22/6,21/6, the fourth term is 2β=1.