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anonymous

  • one year ago

Find the 6th term of a geometric sequence t3 = 444 and t7 = 7104. 7104/444=16

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  1. anonymous
    • one year ago
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    @Michele_Laino

  2. anonymous
    • one year ago
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    16/2=8

  3. Michele_Laino
    • one year ago
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    here we have to find the constant of our sequence

  4. anonymous
    • one year ago
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    okay

  5. Michele_Laino
    • one year ago
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    since your problem asks for 6 terms, and we already have two terms, then we have to add other four terms, between those two terms, namely 444 and 7104. In other words we have to find four numbers, such that: \[\Large 444,\;{a_1},\;{a_2},\;{a_3},\;{a_4},\;7104\] are a geometric sequence

  6. anonymous
    • one year ago
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    okay

  7. Michele_Laino
    • one year ago
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    sorry since t2= 444 and t7= 7104, we have to find only 3 terms not four, so we have this: \[\Large 444,\;{t_4},\;{t_5},\;{t_6},\;7104\]

  8. Michele_Laino
    • one year ago
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    next: using the general formula for the n-th term of a geometric sequence whose constant is q, we can write this: \[\Large 7104 = 444 \cdot {q^4}\] what is q?

  9. Michele_Laino
    • one year ago
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    oops..t3=444

  10. anonymous
    • one year ago
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    q4=16

  11. Michele_Laino
    • one year ago
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    so, q=...?

  12. anonymous
    • one year ago
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    4

  13. Michele_Laino
    • one year ago
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    \[\Large \sqrt[4]{{16}} = ...?\]

  14. anonymous
    • one year ago
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    2

  15. Michele_Laino
    • one year ago
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    correct! so we have: \[\Large {t_4} = 2 \cdot 444 = 888\] and so on...

  16. anonymous
    • one year ago
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    3552

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