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Thread: Pre-calc-arithmetic sequence

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    Pre-calc-arithmetic sequence

    in an arithmetic sequence t sub 2=6 , t sub 6=16 find t sub 12

    sorry i dont know how to do the sub thing.hopefully it makes sense.Thank you!
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    Quote Originally Posted by ep78 View Post
    in an arithmetic sequence t sub 2=6 , t sub 6=16 find t sub 12

    sorry i dont know how to do the sub thing.hopefully it makes sense.Thank you!
    see here to see what the variables mean.

    since the general term is given by $\displaystyle t_n = t_1 + (n - 1)d$, you have

    $\displaystyle t_2 = 6 = t_1 + d$

    and

    $\displaystyle t_6 = 16 = t_1 + 5d$

    you can solve this system for $\displaystyle t_1$ and $\displaystyle d$ and hence find $\displaystyle t_{12} = t_1 + 11d$
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    Quote Originally Posted by ep78 View Post
    in an arithmetic sequence t sub 2=6 , t sub 6=16 find t sub 12

    sorry i dont know how to do the sub thing.hopefully it makes sense.Thank you!
    For an arithmetic sequence...

    $\displaystyle t_n = t_1 + (n - 1)d$.

    You're told that $\displaystyle t_2 = 6$, so

    $\displaystyle t_2 = t_1 + (2 - 1)d$

    $\displaystyle 6 = t_1 + d$

    and you're also told that $\displaystyle t_6 = 16$, so

    $\displaystyle t_6 = t_1 + (6 - 1)d$

    $\displaystyle 16 = t_1 + 5d$.


    You now have 2 equations in 2 unknowns that you can solve simultaneously. Subtract equation 1 from equation 2 and you should find

    $\displaystyle 10 = 4d$

    $\displaystyle d = \frac{5}{2}$.


    Substitute back into equation 1 to get

    $\displaystyle 6 = t_1 + \frac{5}{2}$

    $\displaystyle t_1 = \frac{7}{2}$.


    Therefore $\displaystyle t_n = \frac{7}{2} + (n - 1)\frac{5}{2}$.


    What is $\displaystyle t_{12}$?
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