Question

If the $p^{th},q^{th},r^{th}$ terms of a H.P. be $a,b,c$ respectively, prove that
$(q-r)bc+(r-p)ca+(p-q)ab=0$.

Answer 1

Given that $a,b,c$ are the $p^{th},q^{th}.r^{th}$ terms of a H.P

$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are the $p^{th},q^{th}.r^{th}$ terms of an A.P

Let the A.P have $x$ as the first term and $d$ as the common difference

$\Rightarrow$$\frac{1}{a}=x+(p-1)d$ ----1

$\Rightarrow$$\frac{1}{b}=x+(q-1)d$ ----2

$\Rightarrow$$\frac{1}{c}=x+(r-1)d$ ----3

1-2 we get $\frac{1}{a}-\frac{1}{b}=(p-q)d$

$d=\frac{b-a}{ab(p-q)}$

Similarly

2-3 we get $d=\frac{c-b}{bc(q-r)}$

3-1 we get $d=\frac{a-c}{ac(r-p)}$

$\Rightarrow$$d=\frac{b-a}{ab(p-q)}=\frac{c-b}{bc(q-r)}=\frac{a-c}{ac(r-p)}$

$\Rightarrow$We know that if $x=\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$ then $x=\frac{a+c+e}{b+d+f}$

$\Rightarrow$$d=\frac{b-a+c-b+a-c}{ab(p-q)+bc(q-r)+ac(r-p)}$

$\Rightarrow$$d=\frac{0}{ab(p-q)+bc(q-r)+ac(r-p)}$

$\Rightarrow$$\left(ab(p-q)+bc(q-r)+ac(r-p)\right)d=0$

$\Rightarrow ab(p-q)+bc(q-r)+ac(r-p)=0$