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  1. #1
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    Help

    If a1,a2,...an are in A.P. where ai>0 then value of
    1/(sq root a1+sq root a2) + 1/(sq root a2+sq root a3) +.......... 1/sq root an-1 + sq root an = ??


    I did not knw how to write it in proper form but hope u understand my question
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  2. #2
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    Quote Originally Posted by cnpranav View Post
    If a1,a2,...an are in A.P. where ai>0 then value of
    1/(sq root a1+sq root a2) + 1/(sq root a2+sq root a3) +.......... 1/sq root an-1 + sq root an = ??


    I did not knw how to write it in proper form but hope u understand my question
    Do you mean \frac{1}{\sqrt{a1}+\sqrt{a2}}+ \frac{1}{\sqrt{a2}+ \sqrt{a3}}+ \cdot\cdot\cdot+ \frac{1}{\sqrt{a_{n-1}+ a_n}}?
    (Edited)

    And a1, a2, ..., an are an arithmetic progression?

    I would start by looking at simple examples.
    Last edited by HallsofIvy; April 2nd 2009 at 11:36 AM.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Do you mean \frac{1}{\sqrt{a1}+\sqrt{a2}}+ \frac{1}{\sqrt{a2}+ \sqrt{a3}}+ \cdot\cdot\cdot+ \sqrt{a_{n-1}+ a_n}?

    And a1, a2, ..., an are an arithmetic progression?

    I would start by looking at simple examples.
    the last expression is also in upon i.e. frac{1}/{\sqrt{an-1}+\sqrt{an}}
    Last edited by cnpranav; April 2nd 2009 at 09:16 AM.
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  4. #4
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    Hello, cnpranav!

    If a_1,a_2,a_3\:\hdots\:a_n are in A.P. where a_i>0

    then find the value of: . S \:=\:\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \hdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}}
    Rationalize the fractions . . .

    . . . . . \frac{1}{\sqrt{a_2} + \sqrt{a_1}}\cdot\frac{\sqrt{a_2}-\sqrt{a_1}}{\sqrt{a_2} - \sqrt{a_1}} \:=\: \frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}

    . . . . . \frac{1}{\sqrt{a_3} + \sqrt{a_2}}\cdot\frac{\sqrt{a_3} - \sqrt{a_2}}{\sqrt{a_3} - \sqrt{a_2}} \:=\: \frac{\sqrt{a_3} - \sqrt{a_2}}{a_3-a_2}

    . . . . . . . . . . . \vdots . . . . . . . . . . . . . \vdots

    . . \frac{1}{\sqrt{a_n} + \sqrt{a_{n-1}}}\cdot\frac{\sqrt{a_n} - \sqrt{a_{n-1}}}{\sqrt{a_n} - \sqrt{a_{n-1}}}  \:=\:\frac{\sqrt{a_n} - \sqrt{a_{n-1}}}{a_n - a_{n-1}}



    Since a_1,a_2,a_3,\hdots are in A.P.,
    . . all the denominators are the common difference, d.


    We have: . S \;=\;\frac{a_2-a_1}{d} + \frac{a_3-a_2}{d} + \frac{a_4-a_3}{d} + \hdots + \frac{a_n - a_{n-1}}{d}

    . . which simplifies to: . S \;=\;\frac{a_n - a_1}{d}

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, cnpranav!

    Rationalize the fractions . . .

    . . . . . \frac{1}{\sqrt{a_2} + \sqrt{a_1}}\cdot\frac{\sqrt{a_2}-\sqrt{a_1}}{\sqrt{a_2} - \sqrt{a_1}} \:=\: \frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}

    . . . . . \frac{1}{\sqrt{a_3} + \sqrt{a_2}}\cdot\frac{\sqrt{a_3} - \sqrt{a_2}}{\sqrt{a_3} - \sqrt{a_2}} \:=\: \frac{\sqrt{a_3} - \sqrt{a_2}}{a_3-a_2}

    . . . . . . . . . . . \vdots . . . . . . . . . . . . . \vdots

    . . \frac{1}{\sqrt{a_n} + \sqrt{a_{n-1}}}\cdot\frac{\sqrt{a_n} - \sqrt{a_{n-1}}}{\sqrt{a_n} - \sqrt{a_{n-1}}}  \:=\:\frac{\sqrt{a_n} - \sqrt{a_{n-1}}}{a_n - a_{n-1}}



    Since a_1,a_2,a_3,\hdots are in A.P.,
    . . all the denominators are the common difference, d.


    We have: . S \;=\;\frac{a_2-a_1}{d} + \frac{a_3-a_2}{d} + \frac{a_4-a_3}{d} + \hdots + \frac{a_n - a_{n-1}}{d}

    . . which simplifies to: . S \;=\;\frac{a_n - a_1}{d}

    after rationalizing the numerator was in under root how did it change??
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  6. #6
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    the answer is given in the form n-1/(sq root a1+ sq root an)
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