Question

If $p^{\mathrm{th}}$ term and $q^{\mathrm{th}}$ term of an A. P. are $\frac{1}{qr}$ and $\frac{1}{pr}$ respectively, then r th term of the A. P. =

Answer 1

Step 1: Given information

$p^{\text{th}}\text{\hspace{0.17em}term}=\frac{1}{qr}\text{and}{q}^{\text{th}}\text{\hspace{0.17em}term}=\frac{1}{pr}$

Apply the formula used

The $n^{th}$of an A.P. having $d$ as a common difference and $a$ as its first term is given by:

$a_n=a+(n-1)d$

Step 2: Solving according to question using the above formula :

$\begin{array}{l}a+(p-1)d=\frac{1}{qr}\mathrm{...}\left(1\right)\\ a+(q-1)d=\frac{1}{pr}\mathrm{...}\left(2\right)\end{array}$

Subtracting equation [2] from equation [1] :

$\begin{array}{c}\Rightarrow (p-1)d-(q-1)d=\frac{1}{qr}-\frac{1}{pr}\\ \Rightarrow [(p-1)-(q-1)]d=\frac{p-q}{pqr}\left[\text{Taking LCM in RHS}\right]\\ \Rightarrow (p-1-q+1)d=\frac{p-q}{pqr}\\ \Rightarrow (p-q)d=\frac{p-q}{pqr}\\ \Rightarrow d=\frac{1}{pqr}\end{array}$

Step 3: Substituting $d$ in equation (1)

$\begin{array}{rcl}& \Rightarrow & a+\left(p-1\right)d=\frac{1}{qr}\\ & \Rightarrow & a+\left(p-1\right)\frac{1}{pqr}=\frac{1}{qr}\\ & \Rightarrow & a=\frac{1}{qr}-\frac{(p-1)}{pqr}\\ & \Rightarrow & a=\frac{p-(p-1)}{pqr}\\ & \Rightarrow & a=\frac{1}{pqr}\end{array}$

Thus , $r^{th}$ term = a + (r-1)d

Step 4: Substituting values of $a\text{and}d$

$\begin{array}{rcl}\therefore r^{th}term& =& \frac{1}{pqr}+(r-1)\left(\frac{1}{pqr}\right)\\ & =& \frac{1+r-1}{pqr}\\ & =& \frac{r}{pqr}\\ & =& \frac{1}{pq}\end{array}$

Hence, the $r^{th}\mathrm{term}\mathrm{is}\frac{1}{pq}$.