Question

The value of the determinant $\left|\begin{array}{ccc}bc& ca& ab\\ p& q& r\\ 1& 1& 1\end{array}\right|$, where $a,b$ and $c$ are respectively the $pth,qth$ and $rth$ terms of a $H.P.,$ is

Answer 1

Expand the determinant we get

$\left|\begin{array}{ccc}bc& ca& ab\\ p& q& r\\ 1& 1& 1\end{array}\right|=1(a\times c\times r-a\times b\times q)+1(a\times b\times p-b\times c\times r)+1(b\times c\times q-p\times c\times a)=a\times c(r-p)+a\times b(p-q)+b\times c(q-r)$

Now given $a,bandc$ are $pth,qthandrth$ terms of a $H.P$ then $\frac{1}{a},\frac{1}{b}and\frac{1}{c}$ are in $A.P$

Let x be the first term in A.P and common difference be d.

$T_p=x+(p-1)\times d=\frac{1}{a}..........\left(1\right)T_q=x+(q-1)\times d=\frac{1}{b}..........\left(2\right)T_r=x+(r-1)\times d=\frac{1}{c}..........\left(3\right)$

Subtract (1) and (2) we get

$(p-q)\times d=\frac{1}{a}-\frac{1}{b}(p-q)\times d=\frac{b-a}{ab}⟹(p-q)\times ab=\frac{b-a}{d}..........\left(4\right)$

Subtract (3) and (1) we get

$(r-p)\times ac=\frac{a-c}{d}..........\left(5\right)$

Subtract (2) and (1) we get

$(q-r)\times bc=\frac{c-b}{d}..........\left(6\right)$

Add (4) , (5) and (6) we get

$(p-q)\times ab+(r-p)\times ac+(q-r)\times bc=\frac{b-a+a-c+c-b}{d}=0$

Hence determinant $\left|\begin{array}{ccc}bc& ca& ab\\ p& q& r\\ 1& 1& 1\end{array}\right|=0$