If m th term of an AP is 1/n and n th term is 1/m. Show that the sum of the mn terms is 1/2 (mn+1)

If m th term of an AP is 1/n and n th term is 1/m. Show that the sum o

1.	If m th term of an AP is 1/n and n th term is 1/m. Show that the sum of the mn terms is 1/2 (mn+1)
Answer» Given {tex}m^{th}term=\\frac{1}{n}{/tex}\xa0i.e. {tex}{/tex}\xa0{tex}a_m=\\frac{1}{n}{/tex}{tex}\\Rightarrow (a+(m-1)d)=\\frac{1}{n}{/tex} ......(1)and{tex}n^{th}term=\\frac{1}{m}{/tex}i.e. {tex}a_n=\\frac{1}{m}{/tex}{tex}\\Rightarrow a+(n-1)d=\\frac{1}{m}{/tex}.....(2)subtracting Eq(1) from\xa0Eq(2) , we get\xa0{tex}\\Rightarrow[ a+(n-1)d] -[(a+(m-1)d)] =\\frac{1}{m} - \\frac{1}{n}{/tex}{tex}\\Rightarrow a+(n-1)d -a+(m-1)d =\\frac{n - m}{mn} {/tex}{tex}\\Rightarrow nd- d -md+d =\\frac{n - m}{mn} {/tex}{tex}\\Rightarrow (n -m)d =\\frac{n - m}{mn} {/tex}{tex}d=\\frac{1}{mn}{/tex}\xa0and {tex}a=\\frac{1}{mn}{/tex}Now, sum of mn terms,{tex}{S_{mn}} = \\frac{{mn}}{2}[2a + (mn - 1)d]{/tex}{tex}\\Rightarrow[{S_{mn}} = \\frac{{mn}}{2}[\\frac{2}{{mn}} + (mn - 1)\\frac{1}{{mn}}]{/tex}={tex}\\frac{1}{2}(mn+1){/tex}\xa0Hence Proved.