V

Let a and d be the first term and common difference respectively of the given A.P. Thenan = a + (n – 1)d{tex}\\frac { 1 } { n } ={/tex}\xa0mth term\xa0{tex}\\Rightarrow \\frac { 1 } { n } {/tex}= a + ( m – 1 ) d\xa0…(i){tex}\\frac { 1 } { m }{/tex}= nth term{tex}\\Rightarrow \\frac { 1 } { m } {/tex}= a + ( n – 1 ) d\xa0…(ii)On subtracting equation (ii) from equation (i), we get{tex}\\frac { 1 } { n } – \\frac { 1 } { m } = {/tex} [a+ (m-1) d] -[ a+ (n -1)d]= a + md – d – a – nd + d{tex}= ( m – n ) d{/tex}{tex} \\Rightarrow \\frac { m – n } { m n } = ( m – n ) d {/tex}{tex}\\Rightarrow d = \\frac { 1 } { m n }{/tex}Putting d = {tex}\\frac { 1 } { m n }{/tex}\xa0in equation (i), we get{tex}\\frac { 1 } { n } = a + \\frac { ( m – 1 ) } { m n } {/tex}{tex}\\Rightarrow \\frac { 1 } { n } = a + \\frac { 1 } { n } – \\frac { 1 } { m n } {/tex}{tex}\\Rightarrow a = \\frac { 1 } { m n }{/tex}{tex}\\therefore{/tex}\xa0(mn)th term = a + (mn – 1) d=\xa0{tex}\\frac { 1 } { m n } + ( m n – 1 ) \\frac { 1 } { m n } {/tex}{tex}\\left[ \\because a = \\frac { 1 } { m n } = d \\right]{/tex}= {tex}\\frac { 1 } { m n } + \\frac { mn } { m n } – \\frac { 1 } { m n }{/tex}= 1

A+(m-1)d=1/n. And. A + ( n-1)d =1/m……………….An+mnd-dn=1. And. Am + mnd – dm = 1…….equating both these equations………An+mnd-dn=am+mnd-dn…..N(a-d)=m(a-d)…..M=n…….Smn=mn/2×(2a+(mn-1)d)……Smn=mn/2×(2a+mnd-d)…….Smn=
Haa
Sirf ncert chalega
Rd se mt bnao..time waste hoga
also in Rs aggarwal
Rd ka question hai

Consult R.D sharma pg-9.18 example no. 18