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Math Help - Arithmetic Progression

  1. #1
    Member sbhatnagar's Avatar
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    Arithmetic Progression

    I need help with this question involving an A.P.

    The question says: If the sum of n terms of an A.P is cn^2 then the sum of squares of these n terms will be equal to:

    (a) \frac{n(4n^2-1)c^2}{6} (b) \frac{n(4n^2+1)c^2}{3} (c) \frac{n(4n^2-1)c^2}{3} (d) \frac{n(4n^2+1)c^2}{6}


    What I did: Let the sum of n terms of the A.P be S_{n}.

    \Rightarrow S_{n}=\frac{n}{2}[a_{1}+a_{n}]

    \Rightarrow cn^2=\frac{n}{2}[a_{1}+a_{n}]

    \Rightarrow 2cn=a_{1}+a_{n} ...(i)

    When n=1

    \Rightarrow 2c=a_{1}+a_{1}

    \Rightarrow c=a_{1}  ...(ii)

    In equation (i)

    \Rightarrow 2cn-c=a_{n} Using (ii)

    I have found the value of a_{n}. What should I do next?
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  2. #2
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    Re: Arithmetic Progression

    Seems no clue to square the terms just with  a_{1} and a_{n}
    I would use difference method:
    a_{n}=S_{n}-S_{n-1}=c n^{2}-c(n-1)^{2}

    c n^{2}-c(n-1)^{2} <-- square this and sum from 1 to n

    \sum_{r=1}^{n}[c r^{2}-c(r-1)^{2}]^{2} ...works... -->ans (c)

    I'm trying to get used to TEX
    Last edited by BookEnquiry; October 7th 2011 at 06:16 AM.
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  3. #3
    Member sbhatnagar's Avatar
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    Question Re: Arithmetic Progression

    Thanks, I got it.

    a_{n}=2cn-c

    \Rightarrow a_{n}^2=4c^2n^2+c^2-4c^2n

    I just have to find : \sum_{n=1}^{n}a_{n}^2

    \sum_{n=1}^{n}a_{n}^2

    = \sum_{n=1}^{n}(4c^2n^2+c^2-4c^2n)

    =\frac{4c^2n(n+1)(2n+1)}{6}+nc^2-\frac{4c^2n(n+1)}{2}

    =\frac{n(4n^2-1)c^2}{3}

    So the correct answer is (c).

    Am I correct?
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  4. #4
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    Re: Arithmetic Progression

    Sure. My chemistry teacher told us the formula of \sum_{a=1}^{n}a^{2} is quite easy to remember. For e.g. when calculate 1^{2}+2^{2}+3^{2}... +5^{2} just multiply successively 5*6*(5+6)/6 that's it.

    If 1^{2}+2^{2}+3^{2}... +7^{2} just multiply 7*8*(7+8)/6 successively.
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  5. #5
    Member sbhatnagar's Avatar
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    Re: Arithmetic Progression

    yeah. The formula is \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}
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