Q) If the mth term of an AP is 1/n and the nth term is 1/m, then show that the sum of (mn)th term is \frac1 2(mn+1).


Let first term of the given AP = a

Common difference of the given AP = d


a_m=\frac{1}{n}\Rightarrow a+(m-1)d=\frac{1}{n}…………….(I)

a_n=\frac{1}{m}\Rightarrow a+(n-1)d=\frac{1}{m}…………….(ii)

Subtracting equation (ii) from equation (ii),we get


=\cancel a+(m-1)d-\cancel a-(n-1)d=\frac{1}{n}-\frac{1}{m}

=md-\cancel d-nd+\cancel d=\frac{1}{n}-\frac{1}{m}


\Rightarrow (m-n)d=\frac{m-n}{mn}

\Rightarrow \cancel{(m-n)}d=\frac{\cancel{m-n}}{mn}

\Rightarrow d=\frac{1}{mn}

Putting d=\frac{1}{mn} in equation (i), we get

a+(m-1)\frac{1}{mn} =\frac{1}{n}

\Rightarrow a+\cancel m\cdot\frac{1}{\cancel {{m}}{n}}- \frac{1}{mn} =\frac{1}{n}

\Rightarrow a+\cancel{\frac{1}{n}}- \frac{1}{mn} =\cancel{\frac{1}{n}}

\Rightarrow a=\frac{1}{mn}

Next, we  know that the sum of n terms of an AP is given by: S_n = \frac{n}{2} (2a + (n - 1) d)

Therefore the sum of mn terms will be given by: S_{mn} = \frac{mn}{2} (2a + (mn - 1) d)

By substituting values of a and d in the above formula, we get: 

S_{mn} = \frac{mn}{2} (2(\frac{1}{mn})+ (mn - 1) (\frac{1}{mn}))

S_{mn} = \frac{mn}{2}( \frac{2}{mn}+\cancel{mn}\cdot\frac{1}{\cancel{mn}} -  \frac{1}{mn})

S_{mn} = \frac{mn}{2}( \frac{1}{mn}+1)

S_{mn} = \frac{1}{2}( \frac{\cancel{mn}}{\cancel{mn}}+mn)

S_{mn} = \frac{1}{2}(mn+1)

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