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Math Help - Arithmetic Progression- Need help in understanding.

  1. #1
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    Arithmetic Progression- Need help in understanding.

    Question :If a^x = b^y = c^z and b^2 = ac show that 1/x,1/y,1/z  are in A.P?
    The solution which i got here is as follows.(PFA)
    i don't understand how  2/y = 1/x + 1/z says it to be A.P.
    if i am not wrong, a sequence is said to be in AP if the next number in the sequence differs from the previous number by a constant value D. So I'm confused with the solution.Will someone please explain?
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  2. #2
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    Hello, chakravarthiponmudi!

    If a^x = b^y = c^z and b^2 = ac show that \frac{1}{x},\:\frac{1}{y},\:\frac{1}{z} are in A.P.

    I don't understand how \frac{2}{y} \:=\: \frac{1}{x} + \frac{1}{z} says it to be A.P.

    if i am not wrong, a sequence is said to be in AP if the next number in the sequence
    differs from the previous number by a constant value d. . . . . You are right!
    And they are, too!


    If we have: . \begin{array}{cccc}\dfrac{1}{y} - \dfrac{1}{x} &=& d & [1] \\ \\[-3mm]\dfrac{1}{z} - \dfrac{1}{y} &=& d & [2] \end{array}

    Equate [1] and [2]: . \frac{1}{y} - \frac{1}{x} \:=\:\frac{1}{z} - \frac{1}{y} \quad\Rightarrow\quad \frac{2}{y} \:=\:\frac{1}{x} + \frac{1}{z}\quad\hdots see?

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  3. #3
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    thanks

    Thank you!
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