Find the common ratio and an explicit form in each of the following geometric sequences.c.4 / 3,2 / 3,1 / 3,1 / 6 , ...

Accepted Solution

Answer:The common ratio is [tex] \frac{1}{2}[/tex] and the explicit form is [tex]a_{n} =\frac{4}{3}r ^{n-1}[/tex]  for all n≥1.Step-by-step explanation:The general form of a geometric sequence is    [tex]a_{n}=ar^{n-1}[/tex],   for all  n≥1, where r≠0   is the common ratio and [tex]a[/tex] is the first term  of the sequence.From the problem we know the first term is a= [tex]\frac{4}{3}[/tex],   so we only need to calculate   r.  We  know that [tex]a_{2} =\frac{2}{3}[/tex] is the second  term of the sequence, so  we can replace [tex]a_{2}[/tex] and   a into our formula and find the value of r, so we have  for n=2[tex]a_2=ar^{2-1}[/tex]  [tex]\frac{2}{3} =\frac{4}{3} r[/tex] [tex]r=\frac{\frac{2}{3} }{\frac{4}{3} }=\frac{2}{4}=\frac{1}{2}[/tex] so the explicit form is  [tex]a_{n}=a(\frac{1}{2})^{n-1}[/tex] for all  n≥1, where [tex]a=\frac{4}{3}[/tex]We can check the general formula by substituting the values into the equation.[tex]a_1=(\frac{4}{3} )(\frac{1}{2} ^{1-1})=(\frac{4}{3})(1)=\frac{4}{3} =a[/tex][tex]a_2=(\frac{4}{3} )(\frac{1}{2} )^{2-1}=(\frac{4}{3})(\frac{1}{2} )=\frac{2}{3}[/tex][tex]a_3=(\frac{4}{3} )(\frac{1}{2} )^{3-1}=(\frac{4}{3})(\frac{1}{2}^{2} )=(\frac{4}{3})(\frac{1}{4})=\frac{1}{3}[/tex][tex]a_4=(\frac{4}{3} )(\frac{1}{2} )^{4-1}=(\frac{4}{3})(\frac{1}{2}^{3} )=(\frac{4}{3})(\frac{1}{8})=\frac{1}{6}[/tex]We can conclude that the explicit form of the geometric sequence is [tex]a_{n} =\frac{4}{3}r ^{n-1}[/tex]  for all n≥1, and  the common ratio is  [tex]r=\frac{1}{2}[/tex].