Question

If $p^{th}$ term of an $A.P$ is $\frac{1}{q}$ and $q^{th}$ terms is $\frac{1}{p}$, find its $p{q}^{th}$ term. $(p\ne q)$

Answer 1

Given that,

$p^{th}$ term of an A.P. = $\frac{1}{q}$

$q^{th}$ term of an A.P = $\frac{1}{p}$

We know that

A.P term formula

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

so,

$a_p=a+(p-1)d$

$\frac{1}{q}=a+pd-d$ __(1)

now,

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

$a_q=a+(q-1)d$

$\frac{1}{p}=a+qd-d$ __(2)

from (1) - (2) to,

$\frac{1}{q}-\frac{1}{p}=a+pd-d-a-qd+d$

$pd-qd=\frac{p-q}{pq}$

$d(p-q)=\frac{p-q}{pq}$

$d=\frac{1}{pq}$

put the value of d in equation (1) and

$\frac{1}{q}=a+(p-1)d$

$\frac{1}{q}=a+(p-1)\frac{1}{pq}$

$\frac{1}{q}=a+\frac{p}{pq}-\frac{1}{pq}$

$\frac{1}{q}=a+\frac{1}{q}-\frac{1}{pq}$

$a=a-\frac{1}{pq}$

$a=\frac{1}{pq}$

then term of pq of an A.P

$a_{pq}=a+(pq-1)d$

$a_{pq}=\frac{1}{pq}+(pq-1)\times \frac{1}{pq}$

$a_{pq}=\frac{1}{pq}+\frac{pq}{pq}-\frac{1}{pq}$

$a_{pq}=1$ Hence the $p{q}^{th}$ term = $1$