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Thread: 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 $\displaystyle n$ terms of an A.P is $\displaystyle cn^2$ then the sum of squares of these $\displaystyle n$ terms will be equal to:

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


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

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

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

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

    When n=1

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

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

    In equation (i)

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

    I have found the value of $\displaystyle a_{n}$. 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 $\displaystyle a_{1}$ and $\displaystyle a_{n}$
    I would use difference method:
    $\displaystyle a_{n}=S_{n}-S_{n-1}=c n^{2}-c(n-1)^{2}$

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

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

    I'm trying to get used to TEX
    Last edited by BookEnquiry; Oct 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.

    $\displaystyle a_{n}=2cn-c$

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

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

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

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

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

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

    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 $\displaystyle \sum_{a=1}^{n}a^{2}$ is quite easy to remember. For e.g. when calculate $\displaystyle 1^{2}+2^{2}+3^{2}... +5^{2}$ just multiply successively 5*6*(5+6)/6 that's it.

    If $\displaystyle 1^{2}+2^{2}+3^{2}... +7^{2}$ 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 $\displaystyle \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$
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