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Math Help - The sum of the first three terms of an arithmetic sequence is 12. The sum of their s

  1. #1
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    The sum of the first three terms of an arithmetic sequence is 12. The sum of their s

    Question -
    The sum of the first three terms of an arithmetic sequence is 12. The sum of their squares is 66. Find the fourth term

    I know just by trial an error that the first three terms are 1, 4 and 7. therefore the 4th term is 10.
    Although I am able to figure this out, there IS an algebraic way of determining the values to prove that these ar , in fact, the numbers of the sequence.

    I have started along the lines of
    a + (a+d)^2 + (a+2d)^2 = 66
    a + a +(a+d) + (a+d) + (a+2d) + (a+2d) = 66
    6a+6d = 66

    but im not sure where to go from here. i know im on the right track i just cant figure this out. please help !?
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  2. #2
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    Quote Originally Posted by Iainwb.13 View Post
    Question -
    The sum of the first three terms of an arithmetic sequence is 12. The sum of their squares is 66. Find the fourth term

    I know just by trial an error that the first three terms are 1, 4 and 7. therefore the 4th term is 10.
    Although I am able to figure this out, there IS an algebraic way of determining the values to prove that these ar , in fact, the numbers of the sequence.

    I have started along the lines of
    a + (a+d)^2 + (a+2d)^2 = 66
    a + a +(a+d) + (a+d) + (a+2d) + (a+2d) = 66
    6a+6d = 66

    but im not sure where to go from here. i know im on the right track i just cant figure this out. please help !?
    The terms will be: a, a+d, a+2d, a+3d

    Your first condition says:
    a + (a+d) + (a+2d) = 3a + 3d = 12 \qquad \rightarrow \qquad a + d = 4

    Your second condition says:
    a^2 + (a+d)^2 + (a+2d)^2 = a^2 + a^2 + 2ad + d^2  + a^2 + 4ad + 4d^2 = 3a^2 + 6ad + 5d^2 = 66

    So now you have two equations and two unknowns. Plug in the first equation a+d = 4 into the second equation. Then you can solve for a and d

    Then your fourth term is just a+3d
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  3. #3
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    Hello, Iainwb.13!

    You had a good start . . .


    The sum of the first three terms of an arithmetic sequence is 12.
    The sum of their squares is 66.
    Find the fourth term.. .
    . . . There are two answers.

    The first three terms are: . a,\:a+d,\:a+2d

    Their sum is 12:
    . . a + (a+d) + (a+2d) \:=\:12 \quad\Rightarrow\quad 3a + 3d \:=\:12 \quad\Rightarrow\quad d \:=\:4-a .[1]

    The sum of their squares is 66:
    . . a^2 + (a+d)^2 + (a+2d)^2 \:=\:66 \quad\Rightarrow\quad 3a^2 + 6ad + 5d^2 \:=\:66 .[2]


    Substitute [1] into [2]: . 3a^2 + 6a(4-a) + 5(4-a)^2 \:=\:66

    . . which simplifies to: . a^2 - 8a + 7 \:=\:0

    . . which factors: . (a-1)(a-7) \:=\:0

    . . and has roots: . a \;=\;1,\:7

    Substitute into [1]: . d \;=\;3,\:\text{-}3


    There are two possible sequences . . .

    . . \begin{array}{ccccccc}a = 1,\;d = 3\!: & 1,\:4,\:7  & \text{ The fourth term is 10.} \\<br /> <br />
a = 7,\;d = \text{-}3\!: & 7,\:4,\:1 & \text{ The fourth term is -2.}\end{array}

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  4. #4
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    Thank you soo much .
    I tried doing it by myself before i checked your answers and i did end up with exactly that

    thank you so much, your help was greatl appreciated
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