Question

Given the geometric sequence where a1 = 4 and the common ratio is 3, what is the domain for n?

All integers where n ≥ 1

All integers where > 1

All integers where n ≥ 4

All real number

Answer 1

"n" is the number of terms. So it is a positive integer. The least it can be is 1.

Answer 2

but is it this > or the one with line underneath

Answer 3

1 is included. So \(n \ge 1\)

Answer 4

\(\large r={\color{brown}{ 3}}\qquad
\begin{array}{ccllll}
term&value
\\\hline\\
a_1&4\cdot {\color{brown}{ 3}}\\
a_2& (4\cdot 3)\cdot {\color{brown}{ 3}}\\
a_3&(4\cdot 3\cdot 3)\cdot {\color{brown}{ 3}}\\
a_4&(4\cdot 3\cdot 3\cdot 3)\cdot {\color{brown}{ 3}}\\
....& ....\\
a_n& \square ?
\end{array}\)

where do you think that may end ? what could be "n"?

and also notice it starts off from 4 onwards

Answer 5

anyhow... as you can see, the geometric sequence starts off from 4
onwards and onwards to \(+\infty\)

Answer 6

so n > with line under 4 then?

Answer 7

Yes that would be correct