We have, Aurora is practicing and keeping track of her time to complete each station. The x-coordinate is the station number, and the y-coordinate is the time in minutes since the start of the race that she completed the task. \(\displaystyle{\left({1},{3}\right)},{\left({2},{6}\right)},{\left({3},{12}\right)},{\left({4},{24}\right)}.\)

(A) Here, y-coordinate have values as

\(\displaystyle{a}_{{1}}={3},{a}_{{2}}={6},{a}_{{3}}={12},{a}_{{4}}={24},{a}_{{5}},\ldots\ldots{a}_{{n}}\)

In the above sequence,

\(\displaystyle\frac{{a}_{{2}}}{{a}^{1}}=\frac{6}{{3}}={2}\)

\(\displaystyle\frac{{a}_{{3}}}{{a}_{{4}}}=\frac{12}{{6}}={2}\)

Hence, above sequence data is modeling a geometric sequence because it follow the geometric sequence pattern having same common ratio.

(B) We know that for a geometric sequence, nth term of sequence having first term a and common ratio r is given as \(\displaystyle{a}_{{n}}={a}{r}^{{{n}-{1}}}\) ⋯⋯(1)

From the given data, x-coordinate corresponds to station number and y-coordinate corresponds to time t.

For station5, we need to calculate 5-th term of the sequence given by equation (1).

On substituting \(\displaystyle{a}={3},{r}={2},{\quad\text{and}\quad}{n}={5}\) in equation (1) and using explicit formula to determine the time she will complete station 5 as

\(\displaystyle{a}_{{5}}={\left({3}\right)}\cdot{\left({2}\right)}^{{{5}-{1}}}\)

\(\displaystyle={\left({3}\right)}\cdot{\left({2}\right)}^{4}\)

\(\displaystyle={\left({3}\right)}\cdot{\left({16}\right)}\)

\(\displaystyle={48}\) minutes

Hence, the time she will complete station 5 is 48 minutes.

(C)

For station 9, we need to calculate 9^th term of the sequence given by equation (1).

On substituting \(\displaystyle{a}={3},{r}={2},{\quad\text{and}\quad}{n}={5}\) in equation (1) and using explicit formula to determine the time she will complete station 9 as

\(\displaystyle{a}_{{9}}={\left({3}\right)}\cdot{\left({2}\right)}^{{{9}-{1}}}\)

\(\displaystyle={\left({3}\right)}\cdot{\left({2}\right)}^{8}\)

\(\displaystyle={\left({3}\right)}\cdot{\left({256}\right)}\)

\(\displaystyle={768}\) minutes

Hence, the time she will complete station 9 is 768 minutes.

(A) Here, y-coordinate have values as

\(\displaystyle{a}_{{1}}={3},{a}_{{2}}={6},{a}_{{3}}={12},{a}_{{4}}={24},{a}_{{5}},\ldots\ldots{a}_{{n}}\)

In the above sequence,

\(\displaystyle\frac{{a}_{{2}}}{{a}^{1}}=\frac{6}{{3}}={2}\)

\(\displaystyle\frac{{a}_{{3}}}{{a}_{{4}}}=\frac{12}{{6}}={2}\)

Hence, above sequence data is modeling a geometric sequence because it follow the geometric sequence pattern having same common ratio.

(B) We know that for a geometric sequence, nth term of sequence having first term a and common ratio r is given as \(\displaystyle{a}_{{n}}={a}{r}^{{{n}-{1}}}\) ⋯⋯(1)

From the given data, x-coordinate corresponds to station number and y-coordinate corresponds to time t.

For station5, we need to calculate 5-th term of the sequence given by equation (1).

On substituting \(\displaystyle{a}={3},{r}={2},{\quad\text{and}\quad}{n}={5}\) in equation (1) and using explicit formula to determine the time she will complete station 5 as

\(\displaystyle{a}_{{5}}={\left({3}\right)}\cdot{\left({2}\right)}^{{{5}-{1}}}\)

\(\displaystyle={\left({3}\right)}\cdot{\left({2}\right)}^{4}\)

\(\displaystyle={\left({3}\right)}\cdot{\left({16}\right)}\)

\(\displaystyle={48}\) minutes

Hence, the time she will complete station 5 is 48 minutes.

(C)

For station 9, we need to calculate 9^th term of the sequence given by equation (1).

On substituting \(\displaystyle{a}={3},{r}={2},{\quad\text{and}\quad}{n}={5}\) in equation (1) and using explicit formula to determine the time she will complete station 9 as

\(\displaystyle{a}_{{9}}={\left({3}\right)}\cdot{\left({2}\right)}^{{{9}-{1}}}\)

\(\displaystyle={\left({3}\right)}\cdot{\left({2}\right)}^{8}\)

\(\displaystyle={\left({3}\right)}\cdot{\left({256}\right)}\)

\(\displaystyle={768}\) minutes

Hence, the time she will complete station 9 is 768 minutes.