Question

In an arithmetic progression if $m\left( {{a_m}} \right) = n\left( {{a_n}} \right)$. Find the value of ${\left( {{a_{m + n}}} \right)^{th}}$ term.

Answer 1

Hint: Here, the ${m^{th}}$ and the ${n^{th}}$ term of an arithmetic progression satisfy the relation $m\left( {{a_m}} \right) = n\left( {{a_n}} \right)$. we have to calculate the \[{\left( {m + n} \right)^{th}}\] term of the same arithmetic progression. The ${n^{th}}$ term of an arithmetic progression is given by a formula ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term, $d$ is the common difference and $n$ is the number of term of the given $A.P$.

Complete step-by-step solution:
The given relation between the ${m^{th}}$ and the ${n^{th}}$ terms is $m\left( {{a_m}} \right) = n\left( {{a_n}} \right)$.
We know that the ${n^{th}}$ term of an arithmetic progression is given by ${a_n} = a + \left( {n - 1} \right)d$.
Suppose the first term of an arithmetic progression is $a$ and the common difference is $d$, we have to find the value of the ${n^{th}}$ and the ${m^{th}}$ terms. So, by applying above given formula we get,
${a_m} = a + \left( {m - 1} \right)d$.
${a_n} = a + \left( {n - 1} \right)d$.
Now, putting these values in the given relation, we get,
$
   \Rightarrow m\left\{ {a + \left( {m - 1} \right)d} \right\} = n\left\{ {a + \left( {n - 1} \right)d} \right\} \\
   \Rightarrow ma + m\left( {m - 1} \right)d = na + n\left( {n - 1} \right)d \\
   \Rightarrow ma - na = \left( {{n^2} - n} \right)d - \left( {{m^2} - m} \right)d
 $
Taking $a$ as a common from the terms on the RHS and $d$ as a common from the terms on the left hand side. We get,
$
   \Rightarrow a\left( {m - n} \right) = d\left( {{n^2} - n - {m^2} + m} \right) \\
   \Rightarrow - a\left( {n - m} \right) = d\left( {{n^2} - {m^2} - n + m} \right)
 $
Now, applying the formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$. We get,
\[ \Rightarrow - \left( {n - m} \right) = d\left\{ {\left( {n - m} \right)\left( {n + m} \right) - \left( {n - m} \right)} \right\}\]
Taking $\left( {n - m} \right)$from the terms on the RHS. We get,
$
   \Rightarrow - a\left( {n - m} \right) = d\left( {n - m} \right)\left( {m + n - 1} \right) \\
   \Rightarrow - a = d\left( {m + n - 1} \right)
 $
$ \Rightarrow \therefore a = - d\left( {m + n - 1} \right)$ - - - - - - - - - - - - -(1)
Now, we have to calculate the ${\left( {m + n} \right)^{th}}$ term of the arithmetic progression. By applying above given formula we get,
${a_{m + n}} = a + \left( {m + n - 1} \right)d$.
From the equation (1) we get the value of $a = - d\left( {m + n - 1} \right)$. Putting the value of $a$, we get,
$
   \Rightarrow {a_{m + n}} = - d\left( {m + n - 1} \right) + \left( {m + n - 1} \right)d \\
  \therefore {a_{m + n}} = 0
 $

Thus, we get the value of the ${\left( {m + n} \right)^{th}}$ term of AP is zero.

Note: The summation of the $n$ terms of an arithmetic progression is given by the formula ${S_n} = \dfrac{n}{2}\left\{ {2a + \left( {n - 1} \right)d} \right\}$ where $a$ and $d$ are the first term and the common difference of an AP.