in each of the following geometric sequences, find (i) the 7th term; (ii) the nth term. a. 1/2, 3/4, 9/8, .... b. √5, 5, 5√5, ....c. - 1/2, 3/4, - 9/8, ....

Part a
we have
a. 1/2, 3/4, 9/8, ....
a1=1/2
a2=3/4
a3=9/8
so
a2/a1=(3/4)/(1/2)=3/2
a3/a2=(9/8)/(3/4)=3/2
that means
the common ratio r=3/2
the general formula is
[tex]a_n=a_1\cdot r^{(n-1)}[/tex]substitute given values
[tex]a_n=\frac{1}{2}\cdot(\frac{3}{2})^{(n-1)}[/tex]For n=7
substitute
[tex]\begin{gathered} a_7=\frac{1}{2}\cdot(\frac{3}{2})^{(7-1)} \\ \\ a_7=\frac{1}{2}\cdot\frac{729}{64} \\ \\ a_7=\frac{729}{128} \end{gathered}[/tex]Part b
we have
b. √5, 5, 5√5, ....
a1=√5
a2=5
a3=5√5
so
a2/a1=5/√5=√5
a3/a2=5√5/5=√5
the common ratio is r=√5
the general formula is
[tex]a_n=a_1\cdot r^{(n-1)}[/tex]substitute given values
[tex]a_n=\sqrt[]{5}_{}\cdot(\sqrt[]{5}^{(n-1)})[/tex]For n=7
[tex]\begin{gathered} a_7=\sqrt[]{5}_{}\cdot(\sqrt[]{5}^{(7-1)}) \\ a_7=125\sqrt[]{5}_{} \end{gathered}[/tex]