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 if she

Question

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 if she continued this pattern when was the first time she ran more that 1 mile explain

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Ariana 2 weeks 2021-10-10T07:25:11+00:00 1 Answer 0

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    2021-10-10T07:26:49+00:00

    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 a_n=a_1+(n-1)d, where,

    a_n = nth term of sequence,

    a_1 = 1st term of sequence,

    n =  Number of terms in sequence,

    d = Common difference.

    Let us find common difference of our given sequence as:

    \frac{3}{8}-\frac{3}{16}\Rightarrow \frac{6}{16}-\frac{3}{16}=\frac{3}{16}

    Since Vicki needs to cover more than 1 mile, so we nth term of sequence should be greater than 1.

    1<\frac{3}{16}+(n-1)\cdot \frac{3}{16}

    Let us solve for n.

    1<\frac{3}{16}+\frac{3}{16}n-\frac{3}{16}

    1<\frac{3}{16}n

    1\cdot \frac{16}{3}<\frac{16}{3}\cdot \frac{3}{16}n

    5.333<n

    n>5.333

    We can also write next terms of our sequence as:

    \frac{3}{16},\frac{6}{16}, \frac{9}{16},\frac{12}{16},\frac{15}{16},\frac{18}{16}

    Therefore, Vicki will run more than 1 mile when she is jogging for 6th time.

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