For fixed integers \(n,k \), there exists a sequence of \(n\) terms \(x_1,x_2,x_3,\ldots,x_n\), such that the sum of any \(k\) distinct terms is equal to the sum of the all the other term minus the serial number of each the \(k\) terms. After some calculations, it is found out that each term is in an arithmetic sequence, they have a common difference which is denoted by \(d(n,k)\). Compute \[ d(2016,2015)+d(2015,1729).\]

**Details and Assumptions**

As a explicit example if \(n=4,k=2\). then

\[x_\boxed{1}+x_\boxed{2}=x_3+x_4-\boxed{1}-\boxed{2}\]

**Extra credit**: Solve for general term and also when the terms doesn't start at \(n_1\) but at \(n_a\) for positive integer \(a\).

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