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osanseviero

  • 2 years ago

In a geometric series, a2=6 a5=48 Which is the first term, which is the constant and write the progression

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  1. osanseviero
    • 2 years ago
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    I know that \[an=a1*r ^{n-1}\] What goes next?

  2. osanseviero
    • 2 years ago
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    6=a1*r^(n-1) ?

  3. osanseviero
    • 2 years ago
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    6=a1*r^(2-1), 6=a1*r

  4. Noura11
    • 2 years ago
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    As you said : \[a_n=a_1\times q^{n-1}\] where q is the constant of the serie, so we have : \[a_2=a_1\times q\\a_5=a_1\times q^4\] Is that true ?

  5. osanseviero
    • 2 years ago
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    Yep

  6. Noura11
    • 2 years ago
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    so we get : \[6=a_1\times q~~~~~~(1)\\48=a_1\times q^4~~~(2)\] Now, we can divide the equation (2) over (1) , what should we get ?

  7. osanseviero
    • 2 years ago
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    8=q^3, q=2!

  8. Noura11
    • 2 years ago
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    Good. Now the 1st term can be found easily, can't it ?

  9. osanseviero
    • 2 years ago
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    yepp :D

  10. osanseviero
    • 2 years ago
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    3 :) thanks

  11. osanseviero
    • 2 years ago
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    So an=3*2^n-1 ?

  12. Noura11
    • 2 years ago
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    Yes, it is ! And you are welcome !

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