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Math Help - not working out for me! geometric series

  1. #1
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    Unhappy not working out for me! geometric series

    the third term of a geometric sequence is t3= -75 and the sixth term is t6=9375. determine the first term and the common ratio.


    how do I do this question I am trying really hard but I still don't comprehend this question!
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  2. #2
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    Quote Originally Posted by frogsrcool View Post
    the third term of a geometric sequence is t3= -75 and the sixth term is t6=9375. determine the first term and the common ratio.


    how do I do this question I am trying really hard but I still don't comprehend this question!
    If you write geometric sequences on a graph, they form lines...

    Let us call the geometric term "x"

    Let us call the value of the geometric term "y"

    Now let's find the slope of that line...

    The slope equation is: m=\frac{y_2-y_1}{x_2-x_1}

    Substitute: m=\frac{9375-(-75)}{6-3}

    Subtract: m=\frac{9450}{3}

    Divide: m=3150 (this is a very steep line)

    All lines can be written in the form: y=mx+b

    Substitute: -75=3150(3)+b

    Multiply: -75=9450+b

    Subtract: -9525=b

    Go back to equation: y=mx+b

    Substitute: y=3150x-9525

    Substitute to find the first term: y=3150(1)-9525

    Multiply: y=3150-9525

    Subtract: \boxed{y=-6365}

    but please, CHECK MY ARITHMETIC!!!!!!!!!!!!
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  3. #3
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    Hello, frogsrcool!

    The third term of a geometric sequence is t_3 = -75 and the sixth term is t_6 = 9375
    Determine the first term and the common ratio.

    You're expected to know that the n^{th} term is: . t_n = a\cdot r^{n-1}
    where a is the first term and r is the common ratio.

    The 3^{rd} term is -75: . t_3\:=\:a\cdot r^2 \:=\:-75 [1]

    The 6^{th} term is 9375: . t_6\:=\:a\cdot r^5\:=\:9375 [2]


    Divide [2] by [1]: . \frac{a\cdot r^5}{a\cdot r^2}\:=\:\frac{9375}{-75}\quad\Rightarrow\quad r^3\:=\:-125\quad\Rightarrow\quad\boxed{r = -5}

    Substitute into [1]: . a(-5)^2 \:=\:-75\quad\Rightarrow\quad\boxed{a = -3}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, frogsrcool!


    You're expected to know that the n^{th} term is: . t_n = a\cdot r^{n-1}
    where a is the first term and r is the common ratio.

    The 3^{rd} term is -75: . t_3\:=\:a\cdot r^2 \:=\:-75 [1]

    The 6^{th} term is 9375: . t_6\:=\:a\cdot r^5\:=\:9375 [2]


    Divide [2] by [1]: . \frac{a\cdot r^5}{a\cdot r^2}\:=\:\frac{9375}{-75}\quad\Rightarrow\quad r^3\:=\:-125\quad\Rightarrow\quad\boxed{r = -5}

    Substitute into [1]: . a(-5)^2 \:=\:-75\quad\Rightarrow\quad\boxed{a = -3}

    I thought geometric sequences were linear, and it was algebraic that aren't...

    Or maybe it was arithmetic are linear and geometric aren't...

    Anyway, listen to Soroban before listening to me.
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  5. #5
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    Quote Originally Posted by Quick View Post
    I thought geometric sequences were linear, and it was algebraic that aren't...

    Or maybe it was arithmetic are linear and geometric aren't...

    Anyway, listen to Soroban before listening to me.
    Arithmetic sequences differ by a konstant:
    a,a+k,a+2k,a+3k,...

    Geometric sequences differ by a product:
    a,ak,ak^2,ak^3,...

    Both these sequenes are popular because there is an easy way to sum them.
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