C2 GP

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  1. Ganhad's Avatar
    • Exalted Member
    • Posts: 346
    C2 GP
    Find the number of terms in each of these geometric progressions

    3,6,12,...,768 well i just want to know how to do this then ill know all of em
  2. Killjoy-'s Avatar
    • Exalted and Worshipped Member
    • Posts: 1,083
    Re: C2 GP
    The general form of any term in a GP is ab^{r-1}, r is the term-to-term ratio, 2 in this case, and a must be 3.

    So set 768 to the equation for the general form and solve for n. (n is the term number of the last term in the sequence.)
    Last edited by Killjoy-; 10-05-2012 at 01:46.
  3. Ganhad's Avatar
    • Exalted Member
    • Posts: 346
    Re: C2 GP
    did that its not happening dont know how to get past that
  4. Killjoy-'s Avatar
    • Exalted and Worshipped Member
    • Posts: 1,083
    Re: C2 GP
    So 768=3 \times 2^{n-1}

    You'll need to take logs to base 2. (Or any base for that matter but 2 is simpler.)

    n-1=log_2(256)
  5. raheem94's Avatar
    • TSR Demigod
    • Posts: 5,512
    Re: C2 GP
    (Original post by Ganhad)
    did that its not happening dont know how to get past that
     768=3 \times 2^{n-1}

    Divide both sides by 3,  256 = 2^{n-1} \implies 256 = 2^n \times 2^{-1} \implies 256 = 2^n \times \frac12 \implies 512 = 2^n

    Now take logs.
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