Step 1
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
\(\displaystyle{a}_{{{1}}}={3},\ {a}_{{{2}}}={6},\ {a}_{{{3}}}={12},\ {a}_{{{4}}}={24},\ {a}_{{{5}}}\cdots{a}_{{{n}}}\)
In the above sequence,
\(\displaystyle{\frac{{{a}_{{{2}}}}}{{{a}_{{{1}}}}}}={\frac{{{6}}}{{{3}}}}={2}\)
\(\displaystyle{\frac{{{a}_{{{3}}}}}{{{a}_{{{2}}}}}}={\frac{{{12}}}{{{6}}}}={2}\)
Hence, above sequence data is modeling a geometric sequence because it follow the geometric sequence pattern having same common ratio.
Step 2
(B) We know that for a geometric sequence, \(\displaystyle{n}^{{{t}{h}}}\) term of sequence having first term a and common ratio r is given as
1) \(\displaystyle{a}_{{{n}}}={a}{r}^{{{n}-{1}}}\)
From the given data, x-coordinate corresponds to station number and y-coordinate corresponds to time t.
For station5, we need to calculate \(\displaystyle{5}^{{{t}{h}}}\) term of the sequence given by equation (1).
On substituting \(\displaystyle{a}={3},\ {r}={2},\) and \(\displaystyle{n}={5}\) 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 \(\displaystyle{9}^{{{t}{h}}}\) term of the sequence given by equation (1).
On substituting \(\displaystyle{a}={3},\ {r}={2},\) and \(\displaystyle{n}={5}\) in equation (1) and using explicit formula to determine the time she will complete station 9 as
\(\displaystyle{a}_{{{9}}}={\left({3}\right)}\cdot{\left({2}\right)}^{{{9}-{1}}}\)
\(\displaystyle={\left({3}\right)}\cdot{\left({2}\right)}^{{{2}}}\)
\(\displaystyle={\left({3}\right)}\cdot{\left({256}\right)}\)
\(\displaystyle={768}\) minutes
Hence, the time she will complete station 9 is 768 minutes.
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
\(\displaystyle{a}_{{{1}}}={3},\ {a}_{{{2}}}={6},\ {a}_{{{3}}}={12},\ {a}_{{{4}}}={24},\ {a}_{{{5}}}\cdots{a}_{{{n}}}\)
In the above sequence,
\(\displaystyle{\frac{{{a}_{{{2}}}}}{{{a}_{{{1}}}}}}={\frac{{{6}}}{{{3}}}}={2}\)
\(\displaystyle{\frac{{{a}_{{{3}}}}}{{{a}_{{{2}}}}}}={\frac{{{12}}}{{{6}}}}={2}\)
Hence, above sequence data is modeling a geometric sequence because it follow the geometric sequence pattern having same common ratio.
Step 2
(B) We know that for a geometric sequence, \(\displaystyle{n}^{{{t}{h}}}\) term of sequence having first term a and common ratio r is given as
1) \(\displaystyle{a}_{{{n}}}={a}{r}^{{{n}-{1}}}\)
From the given data, x-coordinate corresponds to station number and y-coordinate corresponds to time t.
For station5, we need to calculate \(\displaystyle{5}^{{{t}{h}}}\) term of the sequence given by equation (1).
On substituting \(\displaystyle{a}={3},\ {r}={2},\) and \(\displaystyle{n}={5}\) 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 \(\displaystyle{9}^{{{t}{h}}}\) term of the sequence given by equation (1).
On substituting \(\displaystyle{a}={3},\ {r}={2},\) and \(\displaystyle{n}={5}\) in equation (1) and using explicit formula to determine the time she will complete station 9 as
\(\displaystyle{a}_{{{9}}}={\left({3}\right)}\cdot{\left({2}\right)}^{{{9}-{1}}}\)
\(\displaystyle={\left({3}\right)}\cdot{\left({2}\right)}^{{{2}}}\)
\(\displaystyle={\left({3}\right)}\cdot{\left({256}\right)}\)
\(\displaystyle={768}\) minutes
Hence, the time she will complete station 9 is 768 minutes.
