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Thread: Find first term and common ratio of Geometric progression

  1. #1
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    Find first term and common ratio of Geometric progression

    Hi All,

    I need a hint for this one. I think I may be missing something here.

    Find the first term and common ratio if,

    $\displaystyle
    33S_5 = S_{10}
    $

    $\displaystyle
    S_{10} = S_9 + 5
    $

    Thanks for your help!
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  2. #2
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    Awetuouncsygg
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    Who are $\displaystyle S_5, S_9, S_{10}$? Write their formula using the first term (let's say $\displaystyle b_1$) and the ration (q), then you'll have a system of equations in two unknowns.
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  3. #3
    Senior Member Sambit's Avatar
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    I suppose $\displaystyle S_5,S_9,S_{10}$ are, respectively, the sum of first 5,9 and 10 terms--right?

    If this is the case and first term=$\displaystyle a$, ratio=$\displaystyle r$, then the sum of the GP series upto first $\displaystyle n$ terms will be

    $\displaystyle S_n=a+ar+ar^2+ar^3+...+ar^{n-1}$

    $\displaystyle =a[1+r+r^2+r^3+....+r^{n-1}]$

    $\displaystyle =a\frac{r^n-1}{r-1}$

    So your first equation reduces to $\displaystyle 33a\frac{r^5-1}{r-1}=a\frac{r^{10}-1}{r-1}$ and second equation reduces to $\displaystyle a\frac{r^{10}-1}{r-1}=a\frac{r^9-1}{r-1}+5$

    Solve these two equations to get $\displaystyle a$ and $\displaystyle r$.
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  4. #4
    Super Member
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    geometric progressions

    Hi mathguy80,
    Sambits first equation can be simplified to produce a quadratic equation in one unknown (r) SUbstitute r in the second to find a.


    bjh
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  5. #5
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    Thanks guys, That clears things up. The quadratic equation was the key. I was a little confused by the large power and thought I wouldn't be able to factorize. Substituting,$\displaystyle r^5 = p$ made it into a simple quadratic to solve!
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