Let the 2010th term be x
Let the 2011th term be y
Let the 2012th term be z

Since A is the sum of the 2010th term and the 2013th term,
the 2013th term is A-x.

Since B is the sum of the 2010th term and the 2014th term,
the 2014th term is B-x.


So we have these five consecutive Fibonacci terms, 
the 2010th, 2011th, 2012th, 2013th, and 2014th:

x, y, z, A-x, B-x

Every Fibonacci term after the 2nd is the sum of the two
preceding terms, so we have these three equations:

(1)       z = x + y
(2)   (A-x) = y + z
(3)   (B-x) = z + (A-x)

From equation (2), which is

A - x = y + z 
      
add x to both sides:

   A = x + y + z 

Subtract z from both sides

A - z = x + y
    
and by the equation (1):

   z = x + y

So A - z = z since both equal to x + y

add z to both sides:

(4)    A = 2z
  
Simplify (3)

   (B-x) = z + (A-x)
   B - x = z + A - x

Add x to both sides:

      B = z + A

Multiply through by 2

     2B = 2z + 2A

From equation (4), substitute A for 2z

     2B = A + 2A

     2B = 3A

Divide both sides by 2A

     =  

     = 

     =  
 
Edwin