If mth tetm of AP is 1 by n and th term of an AP is 1 bym.show that nm

1.	If mth tetm of AP is 1 by n and th term of an AP is 1 bym.show that nmth term is 1
Answer» 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 asTm = a + (m-l)d and Tn =\xa0a + (n-1)d respectively.Now, Tm = {tex}\\frac{1}{n}{/tex}\xa0and Tn = {tex}\\frac{1}{m}{/tex}\xa0(given).{tex}\\therefore{/tex}\xa0a + (m-1)d = {tex}\\frac{1}{n}{/tex}\xa0......................(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}\xa0d =\xa0{tex}\\frac{1}{{mn}}{/tex}Putting d ={tex}\\frac{1}{{mn}}{/tex}\xa0in (i), we geta +\xa0{tex}\\frac{{(m - 1)}}{{mn}}{/tex}\xa0{tex} \\Rightarrow {/tex}\xa0a = {{tex}\\frac{1}{n}{/tex}-{tex}\\frac{{(m - 1)}}{{mn}}{/tex}} =\xa0{tex}\\frac{1}{{mn}}{/tex}Thus, a={tex}\\frac{1}{{mn}}{/tex}\xa0and d={tex}\\frac{1}{{mn}}{/tex}{tex}\\therefore{/tex}Now, in general (mn)th term can be written as Tmn\xa0= 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.

If mth tetm of AP is 1 by n and th term of an AP is 1 bym.show that nmth term is 1