Question

Let $P=(\frac{1}{x_p},p),Q=(\frac{1}{x_q},q)$ and $R=(\frac{1}{x_r},r)$, where $x_k\ne 0$ denotes the $k^{th}$term of an H.P. for $k\in N$. If the area formed by the points $P,Q$ and $R$ is $\lambda pqr$ sq. units, then the value of $\lambda$ is

Answer 1

If $x_k$ is $k^{th}$ term of a H.P., then
$\frac{1}{x_k}$ is $k^{th}$ term of an A.P.

Now, assuming $A$ and $D$ be the first term and common difference for corresponding A.P., we get
$\frac{1}{x_p}=A+(p-1)D\frac{1}{x_q}=A+(q-1)D\frac{1}{x_r}=A+(r-1)D$

Now, the area of the $△PQR$
$=\frac{1}{2}\left|\begin{array}{cccc}{x}_1& x_2& x_3& x_1\\ y_1& y_2& y_3& y_1\end{array}\right|=\frac{1}{2}\left|\begin{array}{cccc}\frac{1}{x_p}& \frac{1}{x_q}& \frac{1}{x_r}& \frac{1}{x_p}\\ p& q& r& p\end{array}\right|=\frac{1}{2}|(\frac{q}{x_p}-\frac{p}{x_q})+(\frac{r}{x_q}-\frac{q}{x_r})+(\frac{p}{x_r}-\frac{r}{x_p})|=\frac{1}{2}\left|\frac{1}{x_p}\right(q-r)+\frac{1}{x_q}(r-p)+\frac{1}{x_r}(p-q\left)\right|=\frac{1}{2}\left|\right\{A+(p-1)D\left\}\right(q-r)+\{A+(q-1)D\left\}\right(r-p)+\{A+(r-1)D\left\}\right(p-q\left)\right|=\frac{1}{2}|A\times \left(0\right)+D\left[\right(p-1\left)\right(q-r)+(q-1\left)\right(r-p)+(r-1\left)\right(p-q\left)\right]|=0\Rightarrow \lambda \left(pqr\right)=0\therefore \lambda =0(\because pqr\ne 0)$