Question

If the mth term of an AP is $\dfrac{1}{n}$ and the nth term is$\dfrac{1}{m}$, show that the sum of mn terms is $\dfrac{1}{2}(mn + 1)$.

Answer 1

Hint:To solve this question we should know the basic concepts and formulas regarding Arithmetic Progressions or AP. The formulas required to solve this question are,
nth term of an AP: ${a_n} = a + (n - 1)d$ and
Sum of n terms of an AP: ${S_n} = \dfrac{n}{2}(2a + (n - 1)d)$

Complete step-by-step answer:
Arithmetic Progression is one of the important sequences from AP, GP (Geometric Progression) and HP (Harmonic Progression). It is a sequence in which two consecutive terms are separated by a common value known as the AP’s common difference.
Let the first term of the given AP be a, and the common difference be d.
It is given that the mth term of an AP is $\dfrac{1}{n}$
$ \Rightarrow {a_m} = a + (m - 1)d$
$ \Rightarrow \dfrac{1}{n} = a + (m - 1)d$ …Eq1
Also, the nth term of an AP is $\dfrac{1}{m}$
$ \Rightarrow {a_n} = a + (n - 1)d$
$ \Rightarrow \dfrac{1}{m} = a + (n - 1)d$ …Eq2

Subtracting Eq1 and Eq2:
$\,\dfrac{1}{n} - \dfrac{1}{m} = [a + (m - 1)d] - [a + (n - 1)d]$
$\begin{gathered}
   \Rightarrow \dfrac{{m - n}}{{mn}} = md - nd \\
   \Rightarrow \dfrac{{m - n}}{{mn}} = (m - n)d \\
   \Rightarrow d = \dfrac{1}{{mn}} \\
\end{gathered} $
Substituting the value of d in Eq1:
$\begin{gathered}
  \dfrac{1}{n} = a + (m - 1)d \\
  Put\,d = \dfrac{1}{{mn}} \\
   \Rightarrow \dfrac{1}{n} = a + (m - 1)\dfrac{1}{{mn}} \\
   \Rightarrow \dfrac{1}{n} = \dfrac{{amn + (m - 1)}}{{mn}} \\
   \Rightarrow m = amn + m - 1 \\
   \Rightarrow 1 = amn \\
   \Rightarrow a = \dfrac{1}{{mn}} \\
\end{gathered} $

We have to find the sum of mn terms of the given AP.
Using the formula, ${S_n} = \dfrac{n}{2}(2a + (n - 1)d)$
${S_{mn}} = \dfrac{{mn}}{2}(2a + (mn - 1)d)$
Putting the values of a and d in this expression:
$\begin{gathered}
  {S_{mn}} = \dfrac{{mn}}{2}(\dfrac{2}{{mn}} + (mn - 1)\dfrac{1}{{mn}}) \\
   = \dfrac{1}{2}(2 + mn - 1) \\
   = \dfrac{1}{2}(mn + 1) \\
\end{gathered} $
Hence it is proved that the sum of mn terms of the given AP $ = \dfrac{1}{2}(mn + 1)$

Note- This question was a direct formula based question in which we evaluated certain conditions to prove the desired result. We had to find the sum of mn terms of the given AP, so our approach was confined to calculating the values of a and d to directly substitute in the formula of the sum of n terms of an AP.