Question

I'll mark brainly answer plzz solve
An even number of A.Ms are inserted between two
numbers whose sum is 13/6. If the sum of means
exceeds their number by 1, what is the number of
means?​

Answer 1

Step-by-step explanation:

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

Answer:

please refer the images.....................

Step-by-step explanation: