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Math Help - Arithmetic Progression X.x

  1. #1
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    Arithmetic Progression X.x

    Thank you for helping! I am soooooooo frustrated with arithmetic and geometric progression because I keep getting stuck at it. sigh.

    Q1: The sum to n terms of a particular series is given by Sn = 17n - 3(n^2). Show that the series is an arithmetic progression and state the value of it's common difference.

    Q2: The sum of the frist 3 terms and that of the last three terms of an arithmetic progression are 9 and 57 respectively. It is known that the sum of the whole sequence is 121. Find the number of terms in this sequence.

    Thank you!
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    Quote Originally Posted by Tangera View Post
    Thank you for helping! I am soooooooo frustrated with arithmetic and geometric progression because I keep getting stuck at it. sigh.

    Q1: The sum to n terms of a particular series is given by Sn = 17n - 3(n^2). Show that the series is an arithmetic progression and state the value of it's common difference.

    Q2: The sum of the frist 3 terms and that of the last three terms of an arithmetic progression are 9 and 57 respectively. It is known that the sum of the whole sequence is 121. Find the number of terms in this sequence.

    Thank you!
    Q1 From the given series: s_{n+1} - s_n = 14 - 6n .... (1)

    But s_{n+1} - s_n = a_1 + nd .... (2)

    Compare (1) and (2) ......
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    Quote Originally Posted by Tangera View Post
    Q2: The sum of the frist 3 terms and that of the last three terms of an arithmetic progression are 9 and 57 respectively. It is known that the sum of the whole sequence is 121. Find the number of terms in this sequence.

    Thank you!
    A brutal approach:

    "The sum of the frist 3 terms .... of an arithmetic progression are 9":

    a_1 + (a_1 + d) + (a_1 + 2d) = 3a_1 + 3d = 3 \Rightarrow a_1 + d = 1 \Rightarrow {\color{red}a_1 = 1 - d} .... (1)


    "The sum of ..... the last three terms of an arithmetic progression are ..... 57":

    (a_1 + [n-3]d) + (a_1 + [n-2]d) + (a_1 + [n-1]d) = 3a_1 + 3nd - 6d = 57 \Rightarrow {\color{red}a_1 + nd - 2d = 19} .... (2)


    "It is known that the sum of the whole sequence is 121":

    121 = \frac{n(2a_1 + [n-1]d)}{2} \Rightarrow {\color{red}242 = 2na_1 + n(n-1)d} .... (3)

    Substitute (1) into (2): d = \frac{18}{n-3} .... (4)

    Substitute (1) and (4) into (3) and solve for n.
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    Thank you very very much for your fast reply! I will continue working on Q1 now...
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    Quote Originally Posted by mr fantastic View Post
    A brutal approach:

    "The sum of the first 3 terms .... of an arithmetic progression are 9":

    a_1 + (a_1 + d) + (a_1 + 2d) = 3a_1 + 3d = 3 \Rightarrow a_1 + d = 1 \Rightarrow {\color{red}a_1 = 1 - d} .... (1)
    I was thinking...shouldn't the equation be 3a + 3d = 9??
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    Quote Originally Posted by mr fantastic View Post
    A brutal approach:

    "The sum of the frist 3 terms .... of an arithmetic progression are 9":

    a_1 + (a_1 + d) + (a_1 + 2d) = 3a_1 + 3d = {\color{blue} 9} \Rightarrow a_1 + d = {\color{blue}3} \Rightarrow {\color{red}a_1 = {\color{blue}3} - d} .... (1)


    "The sum of ..... the last three terms of an arithmetic progression are ..... 57":

    (a_1 + [n-3]d) + (a_1 + [n-2]d) + (a_1 + [n-1]d) = 3a_1 + 3nd - 6d = 57 \Rightarrow {\color{red}a_1 + nd - 2d = 19} .... (2)


    "It is known that the sum of the whole sequence is 121":

    121 = \frac{n(2a_1 + [n-1]d)}{2} \Rightarrow {\color{red}242 = 2na_1 + n(n-1)d} .... (3)

    Substitute (1) into (2): d = \frac{{\color{blue}16}}{n-3} .... (4)

    Substitute (1) and (4) into (3) and solve for n.
    I made one stupid mistake that has had a ripple effect. I've made the corrections in blue.
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  7. #7
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    Quote Originally Posted by Tangera View Post
    I was thinking...shouldn't the equation be 3a + 3d = 9??
    Indeed. I was making the corrections while you spotted and posted this. That's what I get for being too quick! I get n = 11 as the answer by the way.
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    Okay...me too! yay. Thank you!
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  9. #9
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    Hello, Tangera!

    These are challenging problems . . . Thank you!


    Q1) The sum to n terms of a particular series is given by: S_n \:= \:17n - 3n^2
    Show that the series is an A.P. and find its common difference.
    The formula is: . S_n \;=\;\frac{n}{2}\bigg[2a + (n-1)d\bigg]

    We must hammer S_n \:=\:17n - n^2 into this form


    We have: . S_n \;=\;17n - 3n^2 \;=\;\frac{n}{2}\bigg[34 -6n\bigg]

    Add and subtract 6: . S_n \;=\;\frac{n}{2}\bigg[34 - 6n \:{\color{blue}+\: 6 - 6}\bigg] \;=\;\frac{n}{2}\bigg[28 - 6(n-1)\bigg]

    And we have: . S_n \;=\;\frac{n}{2}\bigg[2(14) + (n-1)(\text{-}6)\bigg]

    . . Therefore: . \boxed{a = 14}\;\;\;\boxed{d = -6}

    The AP is: . 14,\,8,\,2,\,\text{-}4,\,\text{-}10,\,\text{-}16, \hdots



    Q2) The sum of the first 3 terms and that of the last three terms of an A.P.
    are 9 and 57, respectively. It is known that the sum of the whole sequence is 121.
    Find the number of terms in this sequence.
    I hope there is a shorter solution!


    The first three terms are: . a,\;a+d,\;a+2d
    Their sum is 9: . 3a + 3d \:=\:9\quad\Rightarrow\quad a + d \:=\:3 \quad\Rightarrow\quad a \:=\:3-d\;\;{\color{blue}[1]}

    The last three terms are: . a + (n-3)d,\;a+(n-2)d,\;a+(n-1)d
    Their sum is 57: . 3a + 3(n-2)d \:=\:57\quad\Rightarrow\quad a + (n-2)d \:=\:19\;\;{\color{blue}[2]}

    The entire sum is 121: . \frac{n}{2}[2a + (n-1)d] \:=\:121\quad\Rightarrow\quad n[2a + (n-1)d \:=\:242\;\;{\color{blue}[3]}


    Substitute [1] into [2]: . (3-d) + (n-2)d \:=\:19\quad\Rightarrow\quad d \:=\:\frac{16}{n-3}\;\;{\color{blue}[4]}

    Substitute [1] into [3]: . n[2(3-d) + (n-1)d] \:=\:242 \quad\Rightarrow\quad 6n + n(n-3)d \:=\:242\;\;{\color{blue}[5]}


    Substitute [4] into [5]: , 6n + n(n-3)\!\cdot\!\frac{16}{n-3} \:=\:242 \quad\Rightarrow\quad 22n \:=\:242

    . . Therefore: . \boxed{n \:=\:11}

    The AP is: . 1,3,5,7,9,11,13,15,17,19,21



    Too fast for me, Mr. F . . . Good job!
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  10. #10
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    Thank you for your help too, Soroban!
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