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Math Help - Hi I Need Urgent Help!!! geometric series

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    Question Hi I Need Urgent Help!!! geometric series

    A finite geomeric sequence has t1 = 0.1024 and t2 = 0.256. How many terms does sequence have, if its middle term has the value 156.25?
    How do I do this question? I am little bit confused!


    t1 = 0.1024 t2 = 0.256
    r=t2 /t1 = 2.5
    t1r^n-1 = 156.25
    (0.1024)(2.5 ) ^n-1= 156.25

    (0.1024)(2.5 ) ^n-1= 156.25/ (0.1024)
    (2.5 ) ^n-1= 1525.878906
    how do I get n by itself??? am I doing this wrong? thanks for your help!
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    Quote Originally Posted by kassandra View Post
    A finite geomeric sequence has t1 = 0.1024 and t2 = 0.256. How many terms does sequence have, if its middle term has the value 156.25?
    How do I do this question? I am little bit confused!


    t1 = 0.1024 t2 = 0.256
    r=t2 /t1 = 2.5
    t1r^n-1 = 156.25
    (0.1024)(2.5 ) ^n-1= 156.25

    (0.1024)(2.5 ) ^n-1= 156.25/ (0.1024)
    (2.5 ) ^n-1= 1525.878906
    how do I get n by itself??? am I doing this wrong? thanks for your help!
    One tiny little gripe first. The line in red is written incorrectly. It should be:
    (2.5)^{n-1} = \frac{156.25}{0.1024}

    But the rest is correct.

    So, the problem is to solve 2.5^{n-1}= 1525.878906

    I'm going to assume you know logarithms, otherwise you can't solve this. The tricky part is how to do a log with a base of 2.5. I'm not even going to try! 2.5 = 5/2 so: (Edit: Come to think of it you can leave it as 2.5. I was going to do something with the fraction that I realized later I didn't need to do.)
    ln \left [ \left ( \frac{5}{2} \right )^{n-1} \right ] = ln(1525.878906) (Or use log base 10, or your favorite base.)

    (n-1) \cdot ln \left ( \frac{5}{2} \right ) = 7.330325855

    n - 1 = \frac{7.330325855}{ln \left ( \frac{5}{2} \right )}

    n = 1 + \frac{7.330325855}{ln \left ( \frac{5}{2} \right )} = 8
    (with a BIG sigh of relief that it's an integer...)

    -Dan
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