We have the sequence: 2, 6, 18, 54...
If the sequence is arithmetic, there must be a common difference between the terms that remains constant.
This is not the case for this sequence.
We can try by seeing if there is a common factor k such that:
[tex]a_n=k\cdot a_{n-1}[/tex]We can do it by:
[tex]\frac{a_2}{a_1}=\frac{6}{2}=3[/tex][tex]\frac{a_3}{a_2}=\frac{18}{6}=3[/tex][tex]\frac{a_4}{a_3}=\frac{54}{18}=3[/tex]There, we have a geometric sequence, with factor k=3:
[tex]a_n=3\cdot a_{n-1}[/tex]We can relate it to the first term as:
[tex]\begin{gathered} a_2=3\cdot a_1 \\ a_3=3\cdot a_2=3\cdot3\cdot a_1=3^2\cdot a_1 \\ a_4=3\cdot a_3=3\cdot3^2\cdot a_1=3^3\cdot a_1 \\ a_n=3^{n-1}\cdot a_1=3^{n-1}_{}\cdot2 \end{gathered}[/tex]For n=12, we have:
[tex]a_{12}=3^{12-1}\cdot2=3^{11}\cdot2=177,147\cdot2=354,294[/tex]The value of a12 is 354,294.
The answer is d) Geometric, 354,294.
