1)Rewrite the table:
70, 49, 34.3, 24.01, 11.807
2) write the quotient of each number by the number before & notice the value:
49/70= 0.7
34.3/49 = 0.7
24.01/34.3 =0.7
16.0807/24.01 = 0.67 ≈0.7
You notice this is a geometric progression with r 0.7
The last term in a GP =ar^(n-1)
a=70; r= 0.7 n-1= number of terms (days -1)
For last term = 1 ==> then 1=70(0.7)^(n-1)
1/70 = (0.7)^(n-1)
log(1/70) =log[(0.7)^(n-1) ==> log1 - log 70 = (n-1) log(0.7) & you will find that it need 11.92 days to equal on & to be less than it will necessitate 12 Days
3) Domain and Range of this function:
Last term = a₁.rⁿ⁻¹. let last term be y==> f(n) = y =70(0.7)ⁿ⁻¹
or f(n) = y = 70(0.7)ⁿ / 0.7==> f(n) = [(0.7)ⁿ ]/ 100.
This is a decreasing exponential function where the coefficient
raised to n is < 1.
The domain is for all n>= 0.
When n→∞, f(n)→0; For n=0==>f(n) =70. So the range of f(n) is:<=70
