Question

A bacteria population triples in number each day . if there are 2,551,500 bacteria on the 7th day , how many bacteria were present on the first day ?

Answer 1

\[2,551,500=x\times 3^6\]

Answer 2

\[x=\frac{2,551,500}{3^6}=3,500\]

Answer 3

first day population is a
a7 is population of 7th day a7=2551500
ratio is r=3
geometric progression sequence
a7=a1 * r^n-1
n=7
2551500=a1 3^6
a1=2551500/729
=3500
population of bacteria on first day is 3500