Question

If $pth,qth$ and $rth$ terms of an H.P. be $a,b,c$ respectively, then $\Delta =\left|\begin{array}{ccc}bc& ca& ab\\ p& q& r\\ 1& 1& 1\end{array}\right|$ is

• A. $-1$
• B. $abc.pqr$
• C. $pa+qb+rc$
• D. $0$

Answer 1

The correct option is D $0$

As $a,b,c$ are $pth,qth,rth$ terms of $H.P.$

Then $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are $pth,qth,rth$ of $A.P.$

Thus

$\frac{1}{a}=A+(p-1)D,\frac{1}{b}=A+(q-1)D$ and $\frac{1}{c}=A+(r-1)D$

Now

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

Applying $R_1\to R_1-(A-D)R_3-D{R}_2$, we get

$\Delta =abc\left|\begin{array}{ccc}0& 0& 0\\ p& q& r\\ 1& 1& 1\end{array}\right|=0$