No idea about a single term?

The general term of an Harmonic progression can be represented as:

${ H }_{ i }=\frac { 1 }{ a+(i-1)d }$

where a is the reciprocal of the first term and d is the common difference.

$\Rightarrow (\frac { { H }_{ i }+{ H }_{ i+1 } }{ { H }_{ i }-{ H }_{ i+1 } } )=\frac { \frac { 1 }{ a+(i-1)d } +\frac { 1 }{ a+id } }{ \frac { 1 }{ a+(i-1)d } -\frac { 1 }{ a+id } } =\frac { \frac { 2a+2id-d }{ (a+(i-1)d)(a+id) } }{ \frac { d }{ (a+(i-1)d)(a+id) } } =\frac { 2a+2id-d }{ d }$

Hence the summation can be represented as:

$\sum _{ i=1 }^{ 1000 }{ { (-1) }^{ i } } \times (\frac { 2a }{ d } +2i-1)$

Let each term of the summation be represented as ${ S }_{ i }$

$S_{ 2n }-S_{ 2n-1 }={ { (-1) }^{ 2n } }\times (\frac { 2a }{ d } +4n-1)-{ { (-1) }^{ 2n-1 } }\times (\frac { 2a }{ d } +4n-3)=2$

Hence for each pair of such consecutive numbers ( 2n and (2n-1 )) the net sum in the summation is $2$ . Since there are $500$ such pairs of numbers from 1 to 1000 $(1,2),(3,4) ......,(999,1000)$ the total value of the summation is $500\times 2=\boxed { 1000 }$