If the pth​ , qth and the rth term of a H.P. be x, y, z respectively, prove that  yz(q-r) + zx(r-p) + xy(p-q) = 0. NO LINK PLEASE

Dear student,
given pth term of H.P is x
$⇒$pth term of A.P is $\frac{1}{x}$
let first term of A.P=A
common difference=D
$⇒$$A+\left(p-1\right)D=\frac{1}{x}$   ..........(1)
similarly,$A+\left(q-1\right)D=\frac{1}{y}$  .........(2)
and $A+\left(r-1\right)D=\frac{1}{z}$   .............(3)
now (1)-(2), we get
$\left(p-q\right)D=\frac{1}{x}-\frac{1}{y}$
or $\left(p-q\right)=\frac{y-x}{\left(xy\right)D}$  ..........(4)
similarly,(2)-(3), we get
$\left(q-r\right)=\frac{z-y}{\left(zy\right)D}$   .........(5)
and (3)-(1),we get
$\left(r-p\right)=\frac{x-z}{\left(zx\right)D}$   ...............(6)
we have to prove,  $yz\left(q-r\right)+zx\left(r-p\right)+xy\left(p-q\right)=0$
substituting the values of (p-q),(q-r) and (r-p) from (4),(5) and (6), we get
L.H.S
$\frac{yz}{D}\left(\frac{z-y}{yz}\right)+\frac{zx}{D}\left(\frac{x-z}{zx}\right)+\frac{xy}{D}\left(\frac{y-x}{xy}\right)$
or $\frac{z-y+x-z+y-x}{D}=0$
hence proved

Enjoy

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