Find the first six terms of the sequence. a1 = 4, an = an-1 + 8

Question

Find the first six terms of the sequence. a1 = 4, an = an-1 + 8

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Raelynn 2 weeks 2021-09-12T07:38:43+00:00 2 Answers 0

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    2021-09-12T07:40:28+00:00

    \bf \begin{array}{llll} term&\stackrel{a_{n-1}+8}{value}\\ \cline{1-2} a_1&4\\ a_2&\stackrel{4+8}{12}\\ a_3&\stackrel{12+8}{20}\\ a_4&\stackrel{20+8}{28}\\ a_5&\stackrel{28+8}{36}\\ a_6&\stackrel{36+8}{44} \end{array}

    0
    2021-09-12T07:40:30+00:00

    Answer:

    The first 6 terms are 4,12,20,28,36,44

    Step-by-step explanation:

    So we have the recursive sequence

    a_n=a_{n-1}+8 \text{ with } a_1=4.

    If you try to dissect what this really means, it becomes easy.  

    Pretend a_n is a term in your sequence.

    Then a_{n-1} is the term right before or something like a_{n+1} means the term right after.

    So it is telling us to find a term all we have to is add eight to the previous term.

    So the second term a_2 is 4+8=12.

    The third term is a_3 is 12+8=20.

    The fourth term is a_4 is 20+8=28.

    The fifth term is a_5 is 28+8=36

    The sixth term is a_6 is 36+8=44.

    Now sometimes it isn’t that easy to see the pattern from the recursive definition of a relation. Sometimes the easiest way is to just plug in. Let’s do a couple of rounds of that just to see what it looks like.

    a_n=a_{n-1}+8 \text{ with } a_1=4.

    a_2=a_1+8=4+8=12

    a_3=a_2+8=12+8=20

    a_4=a_3+8=20+8=28

    a_5=a_4+8=28+8=36

    a_6=a_5+8=36+8=44

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