Is this even possible?

As long as $0<k<n$ , $d(n,k)=-1/2$ : Let $N = {1, 2, ..., n}$ . For any $i < n$ , choose a subset $S$ of $N$ of size $k-1$ , not including either $i$ or $i+1$ , and let $T$ be the elements of $N$ not in either $S$ nor in ${i,i+1}$ (I couldn't figure out the markup for union etc).

Then we have (after moving the "serial numbers" to the left): $x_i + i + \sum_{j \in S} (x_j+j) = x_{i+1} + \sum_{j \in T} x_j$ $x_{i+1} + i+1 + \sum_{j \in S} (x_j+j) = x_{i} + \sum_{j \in T} x_j$

Differencing: $2 (x_i - x_{i+1}) -1 = 0$ ie $d(n,k) = -1/2$ .

In fact, given that the sequence is allowed to select a single $x_i$ twice, the above argument works even for $k=n$ .

According\quad to\quad the\quad question\quad for\quad some\quad k\quad distinct\quad terms\quad { a }_{ q }\quad ,\quad { a }_{ q+1 }....{ a }_{ q+(k-1) }\\ 2\sum _{ i=q }^{ q+(k-1) }{ { a }_{ i } } =\quad \sum _{ i=1 }^{ n }{ { a }_{ i } } -\left( kq\quad +\quad \frac { k(k-1) }{ 2 } \right) \quad \left[ 1 \right] \\ Similarly\quad for\quad k\quad distinct\quad terms\quad { a }_{ q+1 },{ a }_{ q+2 }.......{ a }_{ q+k }\\ 2\sum _{ i=q+1 }^{ q+k }{ { a }_{ i } } =\quad \sum _{ i=1 }^{ n }{ { a }_{ i } } -\left( kq+\frac { k(k+1) }{ 2 } \right) \quad \left[ 2 \right] \\ \\ \left[ 1 \right] -\left[ 2 \right] \Rightarrow \\ 2({ a }_{ q }-{ a }_{ q+k })=k\quad \left[ 3 \right] \\ But\quad { a }_{ 1 },\quad { a }_{ 2 }.....{ a }_{ n }\quad is\quad an\quad A.P\\ Thus\quad { a }_{ q+k }={ a }_{ q }+kd\quad \left[ 4 \right] \\ Subsituting\quad this\quad value\quad in\quad \left[ 3 \right] :\\ -2kd=k\\ \Rightarrow \boxed { d=\frac { -1 }{ 2 } \quad \forall \quad (n,\quad k) } ,\quad where\quad 2\le k<n\\ Thus\quad in\quad general\quad d(n,k)=\frac { -1 }{ 2 } \\ Thus\quad required\quad answer=\quad 2\times \frac { -1 }{ 2 } =\quad \boxed { -1 }