is given that the number of bacteria doubles every hour.
Therefore, the number of bacteria after every hour will form a G.P.
with first term (a=30) and common ratio (r=2)
∴a
3
=ar
2
=(30)(2)
2
=120
Therefore, the number of bacteria at the end of 2nd hour will be 480.
a
5
=ar
4
=(30)(2)
4
=480
and a
n+1
=ar
n
=(30)(2)
n
Thus the number of bacteria at the end of nth hour will be (30)(2)
n
