Question

In an A.P., if ${m^{th}}$ term is $n$ and ${n^{th}}$ term is $m$, show that its ${r^{th}}$ term is $\left( {m + n - r} \right)$.

Answer 1

Hint: First, write the general term of the ${m^{th}}$ and ${n^{th}}$ term of an A.P. It will give two equations. After that subtract them to get the value of the common difference. Then substitute the value in any equation to get the first term. Now, substitute the value of the first term and the common difference in the general formula of ${r^{th}}$ term and do simplification to get the desired result.

Complete step-by-step answer:
We know the general ${p^{th}}$ term for an AP is,
${a_p} = a + \left( {p - 1} \right)d$
Where $a$ is the first term and $d$ is a common difference.
Now, it is given in the question that the ${m^{th}}$ term of an AP is $n$. So, putting $p = m$ in the general term, we get,
$ \Rightarrow a + \left( {m - 1} \right)d = n$ ….. (1)
Again, it is given in the question that the ${n^{th}}$ term of an AP is $m$. So, putting $p = n$ in the general term, we get,
$ \Rightarrow a + \left( {n - 1} \right)d = m$ ….. (2)
Now, subtract equation (1) from (2),
$ \Rightarrow \left[ {a + \left( {n - 1} \right)d} \right] - \left[ {a + \left( {m - 1} \right)d} \right] = m - n$
Open the brackets and multiply the terms,
$ \Rightarrow a + nd - d - a - md + d = m - n$
Simplify the terms,
$ \Rightarrow nd - md = m - n$
Take $d$ common on the left side,
$ \Rightarrow \left( {n - m} \right)d = m - n$
Divide both sides by $\left( {n - m} \right)$,
$ \Rightarrow d = \dfrac{{m - n}}{{n - m}}$
Simplify the terms,
$ \Rightarrow d = - 1$
Now, substitute the value of $d$ in equation (1),
$ \Rightarrow a + \left( {m - 1} \right) \times - 1 = n$
Open bracket and multiply each term,
$ \Rightarrow a - m + 1 = n$
Simplify the terms,
$ \Rightarrow a = m + n - 1$
Thus, in order to get the ${r^{th}}$ term of the AP, we simply need to put the values of the first term and the common difference of the AP in the general term. Thus, in the formula for the general ${r^{th}}$ term,
$ \Rightarrow {a_r} = a + \left( {r - 1} \right)d$
Substitute the values,
$ \Rightarrow {a_r} = \left( {m + n - 1} \right) + \left( {r - 1} \right) \times - 1$
Open brackets and multiply the terms,
$ \Rightarrow {a_r} = m + n - 1 - r + 1$
Simplify the terms,
$ \Rightarrow {a_r} = m + n - r$

Hence, it is proved that the ${r^{th}}$ term of the arithmetic progression is $\left( {m + n - r} \right)$.

Note:
In this solution, two simultaneous equations are formed where the variables are $a$ and $d$, which are solved explicitly. Without finding the values of $a$ and $d$, it is impossible to prove that the ${r^{th}}$ term of the AP is $\left( {m + n - r} \right)$.