Answer:
[tex]f(x)=16\, (\frac{3}{2} )^{x-1}[/tex]
Step-by-step explanation:
Since we are given the common ratio (3/2), all we need to find to define the geometric sequence, is its multiplicative factor ([tex]a[/tex]) that corresponds to the first term of the sequence - remember that all consecutive terms will be generated by multiplying this first value repeatedly by the common ratio (3/2) as shown below:
[tex]f(1)=a\\f(2)=a\,*\,(\frac{3}{2} )\\f(3)=a\,*\,(\frac{3}{2} )^2\\f(4)=a\,*\,(\frac{3}{2} )^3\\f(5)=a\,*\,(\frac{3}{2} )^4[/tex]
Since we are given the information that [tex]f(5)=81[/tex] we can use this to find the value of the first term:
[tex]f(5)=\,a\,*\,(\frac{3}{2} )^4\\81=\,a\,*(\frac{81}{16} )\\a\,*\,81=\,81\.*\,16\\a\,=\,16[/tex]
Notice as well that the first term doesn't contain the common ratio, the second term contains the common ration (3/2) to the power one, the third one contains the common ratio to the power two, the fourth one contains it to the power three, and so forth. So the exponent at which the common ratio appears is always one unit less than the order (x) of the term in question. This concept helps us finalize the expression for the sequence's formula:
[tex]f(x)=16\,*\,(\frac{3}{2} )^{x-1}[/tex]
