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Math Help - Arithmetic Progression Problem...

  1. #1
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    Arithmetic Progression Problem...

    If a_{1},a_{2},a_{3}.........a_{n} are in AP,where a_{1}>0, then the value of the expression
    \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt  {a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\sqrt{  a_{4}}}+........ up to n terms:

    Here are the options:

    1. \frac{1-n}{\sqrt{a_{1}}+\sqrt{a_{n}}}

    2. \frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}

    3. \frac{n-1}{\sqrt{a_{1}}-\sqrt{a_{n}}}

    4. \frac{1-n}{\sqrt{a_{1}}-\sqrt{a_{n}}}
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  2. #2
    MHF Contributor chisigma's Avatar
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    If the sequence a_{k} is an arithmetic progression then…

    a_{k}= a_{1} + (k-1)\cdot a

    Now we arrive to compute the sum in such a way…

    S= \sum_{k=2}^{n} \frac{1}{\sqrt{a_{k}}+\sqrt {a_{k-1}}}= \sum_{k=2}^{n} \frac{\sqrt{a_{k}} - \sqrt {a_{k-1}}}{a_{k} - a_{k-1}}=

    = \frac {1}{a}\cdot (\sqrt{a_{2}} -\sqrt{a_{1}} + \sqrt{a_{3}} - \sqrt{a_{2}} + \dots + \sqrt {a_{n}} - \sqrt{a_{n-1}}) =

    = \frac {1}{a}\cdot (\sqrt {a_{n}} - \sqrt{a_{1}}) = \frac {1}{a}\cdot (\sqrt {a_{1} + (n-1)\cdot a} - \sqrt{a_{1}})

    Kind regards

    \chi \sigma
    Last edited by chisigma; March 23rd 2009 at 02:49 AM.
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  3. #3
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    Quote Originally Posted by siddscool19 View Post
    If a_{1},a_{2},a_{3}.........a_{n} are in AP,where a_{1}>0, then the value of the expression
    \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt  {a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\sqrt{  a_{4}}}+........ up to n terms:

    Here are the options:

    1. \frac{1-n}{\sqrt{a_{1}}+\sqrt{a_{n}}}

    2. \frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}

    3. \frac{n-1}{\sqrt{a_{1}}-\sqrt{a_{n}}}

    4. \frac{1-n}{\sqrt{a_{1}}-\sqrt{a_{n}}}
    Throughout we use
    a_n = a_1 + (n-1)d (*)

    Each addend of,
    \frac{1}{\sqrt{a_{k}}+\sqrt{a_{k+1}}} = \frac{\sqrt{a_{k}}-\sqrt{a_{k+1}}}{(\sqrt{a_{k}}+\sqrt{a_{k+1}})(\sqr  t{a_{k}}-\sqrt{a_{k+1}})}
    =\frac{\sqrt{a_{k}}-\sqrt{a_{k+1}}}{a_k-a_{k+1}}=\frac{\sqrt{a_{k}}-\sqrt{a_{k+1}}}{a_1+(k-1)d+(a_1-kd)}=\frac{\sqrt{a_{k}}-\sqrt{a_{k+1}}}{-d}

    Now each term has a common denominator namely -d. Adding each term up we see the middle terms in the numerator cancel leaving,
    \frac{\sqrt{a_{1}}-\sqrt{a_{n}}}{-d} (**)

    using (*) again we have d = \frac{a_n-a_1}{n-1}.
    Substituting this into (**) we have,
    \frac{(\sqrt{a_{1}}-\sqrt{a_{n}})(1-n)}{a_n-a_1}
    Factoring the denominator we have,
    \frac{(\sqrt{a_{1}}-\sqrt{a_{n}})(1-n)}{(\sqrt{a_n}-\sqrt{a_1})(\sqrt{a_n}+\sqrt{a_1})}
    Leaving,
    \frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}
    which is answer choice 2 (thanks chisigma)
    Last edited by n0083; March 23rd 2009 at 03:38 AM. Reason: missed minus sign
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by chisigma View Post
    If the sequence a_{k} is an arithmetic progression then…

    a_{k}= a_{1} + (k-1)\cdot a

    Now we arrive to compute the sum in such a way…

    S= \sum_{k=2}^{n} \frac{1}{\sqrt{a_{k}}+\sqrt {a_{k-1}}}= \sum_{k=2}^{n} \frac{\sqrt{a_{k}} - \sqrt {a_{k-1}}}{a_{k} - a_{k-1}}=

    = \frac {1}{a}\cdot (\sqrt{a_{2}} -\sqrt{a_{1}} + \sqrt{a_{3}} - \sqrt{a_{2}} + \dots + \sqrt {a_{n}} - \sqrt{a_{n-1}}) =

    = \frac {1}{a}\cdot (\sqrt {a_{n}} - \sqrt{a_{1}}) = \frac {1}{a}\cdot (\sqrt {a_{1} + (n-1)\cdot a} - \sqrt{a_{1}})

    In order to give a complete answer we consider that...

    (\sqrt{a_{n}}- \sqrt{a_{1}}) = \frac{a_{n} - a_{1}}{\sqrt{a_{n}} + \sqrt{a_{1}}}= \frac{(n - 1)\cdot a}{\sqrt{a_{n}} + \sqrt{a_{1}}}

    ... so that is...

    S= \frac{n - 1}{\sqrt{a_{n}} + \sqrt{a_{1}}}

    The correct choice is 2)...

    Kind regards

    \chi \sigma
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  5. #5
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    Hello, siddscool19!

    If a_1,a_2,a_3, \hdots a_n are in AP, where a_1 > 0, find the value of the expression:

    . . S \;=\;\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{  \sqrt{a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\  sqrt{a_{4}}}+ \hdots + \frac{1}{\sqrt{a_n} + \sqrt{a_{n+1}}}


    Here are the options:

    (1)\;\frac{1-n}{\sqrt{a_{1}}+\sqrt{a_{n}}} \qquad (2)\;\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}} \qquad (3)\;\frac{n-1}{\sqrt{a_{1}}-\sqrt{a_{n}}} \qquad (4)\; \frac{1-n}{\sqrt{a_{1}}-\sqrt{a_{n}}}

    I don't agree with any of their answers.
    Did I miscount?

    Let d be the common difference of the AP.


    The k^{th} term is: . \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}


    Multiply top and bottom by: \sqrt{a_{k+1}} - \sqrt{a_k}

    . . \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \cdot\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{\sqrt{a_{k+1}} - \sqrt{a_k}} \;=\;\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{\underbrace{a_{k+1} - a_k}_{\text{This is }d}} \;=\;\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}


    The series becomes:

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


    Most of the terms cancel out and we are left with: . S \;=\;\frac{\sqrt{a_{n+1}} - \sqrt{a_1}}{d}

    \text{Rationalize: }\;S \;=\;\frac{\sqrt{a_{n+1}} - \sqrt{a_1}}{d} \cdot\frac{\sqrt{a_{n+1}} + \sqrt{a_1}}{\sqrt{a_{n+1}} + \sqrt{a_1}} \;=\; \frac{a_{n+1} - a_1}{d\left(\sqrt{a_{n+1}} + \sqrt{a_1}\right)}


    This can be simplified further, with our knowledge of AP's.

    We know that: . a_{n+1} \:=\:a_1 + nd

    So the numerator is: . (a_1 + nd) - a_1 \:=\:nd

    And the fraction becomes: . S \;=\;\frac{nd}{d\left(\sqrt{a_{n+1}} + \sqrt{a_1}\right)}


    Therefore: . S \;=\;\frac{n}{\sqrt{a_{n+1}} + \sqrt{a_1}}

    . . This is closest to their answer-choice (2).

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  6. #6
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    The answer given in my answer key is choice 1.

    I don't know how to solve it. But the answer is 1) choice.
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