Answer: The explicit rule for the geometric sequence is:
an=15 (1/5)^(n-1), for n=1, 2, 3, 4, ...
Solution:
Sequence:
15, 3, 3/5, 3/25
a1=15
a2=3
a3=3/5
a4=3/25
a2/a1=3/15=(3/3)/(15/3)→a2/a1=1/5
a3/a2=(3/5)/3=(3/5)/(3/1)=(3/5)/(1/3)→a3/a2=1/5
a4/a3=(3/25)/(3/5)=(3/25)(5/3)=5/25=(5/5)/(25/5)→a4/a3=1/5
(an)/(an-1)=(1/5)
Solving for an:
an=(1/5) (an-1)
a1=15
n=2→a2=(1/5)(a2-1)=(1/5)(a1)=(1/5)(15)→a2=15 (1/5)
n=3→a3=(1/5)(a3-1)=(1/5)(a2)=(1/5)[(15) (1/5)]=15 (1/5)^(1+1)→
a3=15 (1/5)^2
n=4→a4=(1/5)(a4-1)=(1/5)(a3)=(1/5)[15 (1/5)^2]=15 (1/5)^(2+1)→
a4=15 (1/5)^3
a1=15→a1=15 (1/5)^0=15 (1/5)^(1-1)
a2=15 (1/5)→a2=15 (1/5)^1=15 (1/5)^(2-1)
a3=15 (1/5)^2=15 (1/5)^(3-1)
a4=15 (1/5)^3=15 (1/5)^(4-1)
Then:
an=15 (1/5)^(n-1), for n=1, 2, 3, 4, ...
