If mth tetm of AP is …

Let a be the first term and d be the common difference of the given AP. Now, we know that in general mth and nth terms of the given A.P can be written as
T_m = a + (m-l)d and T_{n =} a + (n-1)d respectively.
Now, T_m = {tex}\frac{1}{n}{/tex} and T_n = {tex}\frac{1}{m}{/tex} (given).
{tex}\therefore{/tex} a + (m-1)d = {tex}\frac{1}{n}{/tex} ......................(i)
and a + (n-1)d ={tex}\frac{1}{m}{/tex}................ ... (ii)
On subtracting (ii) from (i), we get
(m-n)d = ({tex}\frac{1}{n}{/tex}-{tex}\frac{1}{m}{/tex}) = ({tex}\frac{{m - n}}{{mn}}{/tex}) {tex}\Rightarrow{/tex} d = {tex}\frac{1}{{mn}}{/tex}
Putting d ={tex}\frac{1}{{mn}}{/tex} in (i), we get
a + {tex}\frac{{(m - 1)}}{{mn}}{/tex} {tex} \Rightarrow {/tex} a = {{tex}\frac{1}{n}{/tex}-{tex}\frac{{(m - 1)}}{{mn}}{/tex}} = {tex}\frac{1}{{mn}}{/tex}
Thus, a={tex}\frac{1}{{mn}}{/tex} and d={tex}\frac{1}{{mn}}{/tex}
{tex}\therefore{/tex}Now, in general  (mn)th term can be written as T_mn = a +(mn-1)d
= {{tex}\frac{1}{{mn}}{/tex}+{tex}\frac{{(mn - 1)}}{{mn}}{/tex}} [{tex}\because {/tex}a={tex}\frac{1}{{mn}}{/tex}]
= 1.
Hence, the (mn)th term of the given AP is 1.