Question

In an HP, $pth$ term is $q$ and $qth$ term is $p$. Then $pqth$ term is:

• A. $0$
• B. $1$
• C. $pq$
• D. $pq(p+q)$

Answer 1

The correct option is B

$1$

Explanation for the correct option.

Step 1: Apply the formula for the $nth$ term of HP.

Let the first term of AP be $a$ and the common difference be $d$. We know that the $nth$ term of HP is $t_n=\frac{1}{a+\left(n-1\right)d}$. So,

$$\begin{gathered} t_p=\frac{1}{a+\left(p-1\right)d} \\ \Rightarrow q=\frac{1}{a+\left(p-1\right)d} \\ \Rightarrow \frac{1}{q}=a+\left(p-1\right)d...\left(1\right) \end{gathered}$$

$$\begin{gathered} t_q=\frac{1}{a+\left(q-1\right)d} \\ \Rightarrow p=\frac{1}{a+\left(q-1\right)d} \\ \Rightarrow \frac{1}{p}=a+\left(q-1\right)d...\left(2\right) \end{gathered}$$

Step 2: Find $a\mathrm{and}d$.

Subtracting $\left(2\right)$ from $\left(1\right)$, we get

$\begin{array}{rcl}\frac{1}{q}-\frac{1}{p}& =& \left(p-q\right)d\\ & \Rightarrow & \frac{p-q}{pq}=\left(p-q\right)d\\ \Rightarrow d& =& \frac{1}{pq}...\left(3\right)\end{array}$

By substituting the value of $d$ in $\left(1\right)$, we get

$\begin{array}{rcl}\frac{1}{q}& =& a+\left(p-1\right)\frac{1}{pq}\\ & \Rightarrow & \frac{1}{q}=a+\frac{1}{q}-\frac{1}{pq}\\ \Rightarrow a& =& \frac{1}{pq}...\left(4\right)\end{array}$

Step 3: Find the $pqth$ term of HP.

$\begin{array}{rcl}{t}_{pq}& =& \frac{1}{a+\left(pq-1\right)d}\\ & =& \frac{1}{{\displaystyle \frac{1}{pq}}+\left(pq-1\right){\displaystyle \frac{1}{pq}}}\left[\mathrm{from}\left(3\right)\mathrm{and}\left(4\right)\right]\\ & =& \frac{1}{{\displaystyle \frac{1}{pq}}+1-{\displaystyle \frac{1}{pq}}}\\ & =& 1\end{array}$

Hence, option B is correct.