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We have, `H=(2xy)/(x+y)` <br> `therefore (H)/(x)=(2xy)/(x+y) " and "(H)/(y)=(2x)/(x+y)` <br> By componendo and dividendo, we have <br> `(H+x)/(H-x)=(2y+x+y)/(2y-x-y)=(x+3y)/(y-x)` <br> and `(H+y)/(H-y)=(2x+x+y)/(2x-x-y)=(3x+y)/(x-y)` <br> ` therefore (H+x)/(H-x)+(H+y)/(H-y)=(x+3y)/(y-x)+(3x+y)/(x-y)` <br> ` = (x+3y-3x-y)/(y-x)=(2(y-x))/(y-x)=2` <br> Aliter `(H+x)/(H-x)+(H+y)/(H-y)=2` <br> ` implies ((H+x)/(H-x)-1)=(1-(H+y)/(H-y)) implies (2x)/(H-x)=(-2y)/(H-y)` <br> `i.e.Hx-xy=-Hy+xy implies H(x+y)=2xy` <br> `i.e.H =(2xy)/(x+y)` <br> which is true as, `x,H,y` are in HP. Hence, the required result.**Representation of sequences and different types of series**

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