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Thread: Geometric Progression Question

  1. #1
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    Exclamation Geometric Progression Question

    Question :
    If a,b,c are in Geometric Progression , prove that (1/a+b)+(1/b+c) = (1/b) .

    I found that b^2=ac , and thus c = (b^2)/a and a = (b^2)/c here .
    I managed to solve this problem , by substituting "c" in the equation as , (b^2)/a , and simplifying , after which the left hand side finally equated to 1/b.

    I ran into a doubt , why do I need to substitute only for "c" in the equation to equate ? Why cant one substitute for both "a" , "c" or "a" , "c" , "b" simultaneously ?
    I found that , once I substitute for more than once variable in the equation , the equation no longer equates . This maybe a stupid doubt , but please help me clear that abstract gap in my mind .

    Thanks so much .
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  2. #2
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    Perhaps an alternative method would clear up the confusion. Since they are a geometric they differ by a multiplicative constant, lets call it $\displaystyle r$. That is,

    $\displaystyle b = ar, \qquad c = br = ar^2$

    So $\displaystyle a+b = a(1+r)$ and $\displaystyle b+c = ar(1+r),$

    therefore

    $\displaystyle
    \begin{array}{rcl}
    \dfrac{1}{a+b} + \dfrac{1}{b+c} & = & \dfrac{1}{a(1+r)} + \dfrac{1}{ar(1+r)} \\
    & = & \dfrac{r}{ar(1+r)} + \dfrac{1}{ar(1+r)} \\
    & = & \dfrac{1 + r}{ar(1+r)} \\
    & = & \dfrac{1}{ar} \\
    & = & \dfrac{1}{b}.
    \end{array}
    $

    Hope this helps.
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  3. #3
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    Thanks Rincewind , that really helped me figure it out.

    Can anyone post and explain the derivation for 'n' terms of a Geometric Progression as well ? ( And if possible , for infinite terms too ? )

    Thanks .
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  4. #4
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    Hello, dagamer!

    Can anyone explain the derivation for $\displaystyle n$ terms of a GP?
    (And, if possible, for infinite terms, too?)

    . . .We have: . .$\displaystyle S \;=\; a + ar + ar^2 + ar^3 + \hdots + ar^{n-1}$. . . . .[1]

    Multiply by $\displaystyle r\!:\;\;rS \;=\; \quad\;\; ar + ar^2 + ar^3 + \hdots + ar^{n-1} + ar^n $ .[2]


    Subtract [2] from [1]: .$\displaystyle S - rS \;=\;a - ar^n \quad\Rightarrow\quad S(1-r) \;=\;a(1-r^n)$


    . . Therefore: .$\displaystyle \boxed{S \;=\;a\,\frac{1-r^n}{1-r}}$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    For an infinite series . . .

    . . .We have: . .$\displaystyle S \;=\; a + ar + ar^2 + ar^3 + \hdots$ . [1]

    Multiply by $\displaystyle r\!:\;\;rS \;=\; \quad\;\; ar + ar^2 + ar^3 + \hdots$ . .[2]


    Subtract [2] from [1]: .$\displaystyle S - rS \;=\;a \quad\Rightarrow\quad S(1-r) \:=\:a$


    . . Therefore: .$\displaystyle \boxed{S \;=\;\frac{a}{1-r}}\;\;\text{ for }|r| < 1$

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