Answer:
Jogging 6th time.
Step-by-step explanation:
We have been given that Vicki started jogging the first time she ran she ran 3/16 mile the second time she ran 3/8 mile and the third time she ran 9/16 mile.
We can see that the distance Vicki covers each time forms a arithmetic sequence, where 1st term is 3/16.
We know that an arithmetic sequence is in form [tex]a_n=a_1+(n-1)d[/tex], where,
[tex]a_n[/tex] = nth term of sequence,
[tex]a_1[/tex] = 1st term of sequence,
n = Number of terms in sequence,
d = Common difference.
Let us find common difference of our given sequence as:
[tex]\frac{3}{8}-\frac{3}{16}\Rightarrow \frac{6}{16}-\frac{3}{16}=\frac{3}{16}[/tex]
Since Vicki needs to cover more than 1 mile, so we nth term of sequence should be greater than 1.
[tex]1<\frac{3}{16}+(n-1)\cdot \frac{3}{16}[/tex]
Let us solve for n.
[tex]1<\frac{3}{16}+\frac{3}{16}n-\frac{3}{16}[/tex]
[tex]1<\frac{3}{16}n[/tex]
[tex]1\cdot \frac{16}{3}<\frac{16}{3}\cdot \frac{3}{16}n[/tex]
[tex]5.333<n[/tex]
[tex]n>5.333[/tex]
We can also write next terms of our sequence as:
[tex]\frac{3}{16},\frac{6}{16}, \frac{9}{16},\frac{12}{16},\frac{15}{16},\frac{18}{16}[/tex]
Therefore, Vicki will run more than 1 mile when she is jogging for 6th time.
