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Thread: Simple Arithmetic Series Question

  1. #1
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    Simple Arithmetic Series Question

    In an arithmetic series the 20th term is 14 and the 40th term is -6. Find the 10th term.

    How do I do this?
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    Quote Originally Posted by Viral View Post
    In an arithmetic series the 20th term is 14 and the 40th term is -6. Find the 10th term.

    How do I do this?
    $\displaystyle U_n = a+(n-1)d$

    Where

    • $\displaystyle U_n$ is the nth term
    • $\displaystyle a$ is the first term
    • $\displaystyle n$ is the number of terms
    • $\displaystyle d$ is the common difference.


    $\displaystyle U_{20} = a + (20-1)d = 14$ (eq1)

    $\displaystyle U_{40} = a + (40-1)d = -6$ (eq2)

    Solve simultaneously. It would be easiest to eliminate a first
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    Yeah, I've done that, and got $\displaystyle d = 1 \therefore a = -5$.

    However, when I do:

    $\displaystyle U_{10}=-5+9d=14=v$
    $\displaystyle v=-5+9=4$

    But the answer is supposed to be 24 >< .
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    Quote Originally Posted by Viral View Post
    Yeah, I've done that, and got $\displaystyle d = 1 \therefore a = -5$.

    However, when I do:

    $\displaystyle U_{10}=-5+9d=14=v$
    $\displaystyle v=-5+9=4$

    But the answer is supposed to be 24 >< .
    $\displaystyle U_{20} > U_{40}$ which means $\displaystyle d<0$

    I took equation 2 from equation 1:

    $\displaystyle (a+19d) - (a+39d) = 14 - (-6)$

    $\displaystyle -20d = 20 \: \therefore \: d = -1$


    $\displaystyle a+19(-1) = 14 \: \therefore \: a = 33$



    $\displaystyle U_{10} = a + 9d = 33-9 = 24$
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  5. #5
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    I'm now told: The first three terms of an arithmetic series are 5x, 20 and 3x. Find the values of x and hence the values of the three terms.

    What are the steps to do this?
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    Quote Originally Posted by Viral View Post
    I'm now told: The first three terms of an arithmetic series are 5x, 20 and 3x. Find the values of x and hence the values of the three terms.

    What are the steps to do this?
    By definition: $\displaystyle a = 5x$

    $\displaystyle U_2 = 20 = a+d $

    $\displaystyle U_3 = 3x = a+2d$

    You can then solve U_2 and U_3 simultaneously. Eliminate d since we don't need to find it

    I get $\displaystyle x=5$ and so $\displaystyle a = 25$, $\displaystyle U_2 = 20$ and $\displaystyle U_3 = 15$
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  7. #7
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    Ok, this one is worded really weirdly >< . For this can you just tell me what it's asking, and I'll go from there.

    For which values of x would the expression $\displaystyle -8, x^{2} and 17x$ form the first three terms of an arithmetic series.
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  8. #8
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    Quote Originally Posted by Viral View Post
    Ok, this one is worded really weirdly >< . For this can you just tell me what it's asking, and I'll go from there.

    For which values of x would the expression $\displaystyle -8, x^{2} and 17x$ form the first three terms of an arithmetic series.
    Do you get what I did above with the first three terms? In this case set the numbers you're given here

    $\displaystyle U_1 = a = -8$

    $\displaystyle U_2 = a+d = -8+d = x^2$

    $\displaystyle U_3 = a+2d = -8+2d = 17x$
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