Question

If $pth$ term, $qth$ term and $rth$ term of H.P. are $a,b$ and $c$ respectively.
Then prove that
$bc(q-r)+ca(r-p)+ab(p-q)=0$

Answer 1

$n^{th}term=T_n=\frac{1}{a+(n-1)d}{T}_p=a⟹\frac{1}{a_1+(p-1)d}=a⟹a[a_1+(p-1\left)d\right]=1⟹a_1+(p-1)d=\frac{1}{a}⟹(p-1)d=\frac{1}{a}-a_1⟹(p-1)d=\frac{1-a{a}_1}{a}⟹(p-1)=\frac{1-a{a}_1}{ad}⟹p=\frac{1-a{a}_1}{ad}+1-------\left(1\right)T_q=b⟹\frac{1}{a_1+(q-1)d}=b⟹b[a_1+(q-1\left)d\right]=1⟹a_1+(q-1)d=\frac{1}{b}⟹(q-1)d=\frac{1}{b}-a_1⟹(q-1)d=\frac{1-b{a}_1}{b}⟹(q-1)=\frac{1-b{a}_1}{bd}⟹q=\frac{1-a{a}_1}{bd}+1-------\left(2\right)T_r=c⟹\frac{1}{a_1+(r-1)d}=c⟹c[a_1+(r-1\left)d\right]=1⟹a_1+(r-1)d=\frac{1}{c}⟹(r-1)d=\frac{1}{c}-a_1⟹(r-1)d=\frac{1-c{a}_1}{c}⟹(r-1)=\frac{1-c{a}_1}{cd}⟹r=\frac{1-c{a}_1}{cd}+1-------\left(3\right)$

On solving (1), (2) and (3), we get:

$bc(q-r)+ca(r-p)+ab(p-q)=0$