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If the ${m^{th}}$ term of an A.P is $\dfrac{1}{n}$ and the ${n^{th}}$ term is $\dfrac{1}{m}$ , show that the sum of terms is$\dfrac{1}{2}\left( {nm + 1} \right)$.

Last updated date: 20th Jul 2024
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Hint: Whenever it is given that the terms are in A.P, try to use the formulas of A.P, it can be the formula to find the first term or the last term or the sum which makes the question easier to solve as we obtain some equations by putting the values from the given question.

Let $a$ be the first term and $d$ be the common difference of the given A.P.
Now to find the sum we are going to find the general form of m terms and n terms.
${a_m} = \dfrac{1}{n} \Rightarrow a + \left( {m - 1} \right)d = \dfrac{1}{n}$ $......\left( 1 \right)$
\[{a_n} = \dfrac{1}{m} \Rightarrow a + \left( {n - 1} \right)d = \dfrac{1}{m}\] \[......\left( 2 \right)\]
Now we are going to subtract \[\left( 2 \right)\] from\[\left( 1 \right)\], we get,
\[\left( {m - n} \right)d = \dfrac{1}{n} - \dfrac{1}{m} \Rightarrow \left( {m - n} \right)d = \dfrac{{m - n}}{{mn}} \Rightarrow d = \dfrac{1}{{mn}}\]
The next step is to equate \[d = \dfrac{1}{{mn}}\] in\[\left( 1 \right)\], the equation becomes,
\[a + \left( {m - 1} \right)\dfrac{1}{{mn}} = \dfrac{1}{n} \Rightarrow a + \dfrac{1}{n} - \dfrac{1}{{mn}} = \dfrac{1}{n} \Rightarrow a = \dfrac{1}{{mn}}\]
To find the sum of an A.P of \[r\] terms with \[a\]as the first term and \[d\] as the common difference the formula is:
\[{S_r} = \dfrac{r}{2}\left\{ {2a + \left( {n - 1} \right)d} \right\}\]
Therefore, if we substitute the value of $a$and $d$ in the formula of sum of A.P., we get,
\[{S_{mn}} = \dfrac{{mn}}{2}\left\{ {2a + \left( {mn - 1} \right)d} \right\}\]
Note: In this question, they have asked us to find the sum of total terms, therefore we are going to take\[r\] as $mn$ here.
In the next step we equate the value of,
\[{S_{mn}} = \dfrac{{mn}}{2}\left\{ {\dfrac{2}{{mn}} + \left( {mn - 1} \right) \times \dfrac{1}{{mn}}} \right\} = \dfrac{1}{2}\left( {mn + 1} \right)\]
Hence Proved!