Question

the sum of two numbers is 13/6,and an even number of arithematic means are inserted between them such that their sum exceeds their number by 1.find the number of means inserted.

Answer 1

   given,  X + Y = 13 / 6      --- -(1)

2 k is even number,  where k is a natural number.

There are 2k number of Arithmetic means between X and Y.
 So  X,  X+d, X + 2d ,.... , X+ 2k d,  Y      are in Arithmetic series.

  X + (2k+1) d = Y
  X + (2K+1) d = 13/6 - X
   2 X + (2k+1) d = 13/6    --- (2)

   sum of arithmetic means :  X+d , X+2d,.... X+2k d
       =   [ X+ d +  X + 2k d ] 2k / 2
       =   [ 2 X + (2k+1) d ] k  =  2k + 1  given,  as their number plus 1.

       => 13/6 *  k =  2k + 1      using (2)
       =>  k / 6  = 1 
       =>  k = 6
  The number of means inserted are 2k = 12.

   Using (2), we get ,    2 X + 13 d = 13/6

                           d =  1/6 - 2 X /13

   We can have many number of such series..  Depending on the first number, the common difference of the series can be calculated.

Answer 2

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