C2 GP

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    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
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    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.)
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    did that its not happening dont know how to get past that
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    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)
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    (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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