Permutations And Combinations

  • Question
    CBSEENMA11013136

    The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a GP. Find the numbers.

    Solution

    Let a - d, a, a + d be three numbers in A.P. Since their sum is 21
    ∴    a - d + a + a + d = 21
    rightwards double arrow                        3a = 21    rightwards double arrow a = 7
    ∴   Numbers are 7 - d, 7, 7 + d
    It is given that if the second number is reduced by 1 and third is increased by 1 then resulting three numbers form G.P.
    ∴  7 - d, 7 - 1, 7 + d + 1 are in G.P. or 7 - d, 6, 8 + d are in G.P.

    rightwards double arrow               fraction numerator 6 over denominator 7 minus straight d end fraction space equals space fraction numerator 8 plus straight d over denominator 6 end fraction    rightwards double arrow     (8 + d) (7 - d) = 36
    rightwards double arrow     56 - 8d + 7d - d2 = 36         rightwards double arrow      <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>

    rightwards double arrow                 d = fraction numerator negative 1 plus-or-minus square root of 1 plus 80 end root over denominator 2 end fraction space equals space fraction numerator negative 1 plus-or-minus 9 over denominator 2 end fraction space equals space 4 comma space minus space 5
    When a = 7 and d = 4, numbers are 7 - 4,  7 + 4 or 3, 7, 11
    when a = 7 and d = -5, numbers are 7 - (-5), 7, 7 - 5  or 12, 7, 2

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