Question

Let ${T_r}$ be the ${r^{{\text{th}}}}$ term of an A.P., for $r = 1,2,3,....$. If for some positive integers $m,n$ we have ${T_m} = \dfrac{1}{n}$ and ${T_n} = \dfrac{1}{m}$ then ${T_{mn}}$ equals?
(A) $\dfrac{1}{{mn}}$
(B) $\dfrac{1}{m} + \dfrac{1}{n}$
(C) 1
(D) 0

Answer 1

Hint: Use the formula for general term of an A.P. i.e. ${T_n} = a + \left( {n - 1} \right)d$ to frame the equations for ${T_m}$ and ${T_n}$. Find the values of $a$ and $d$ solving these two equations. Put these values in the expression of ${T_{mn}}$ to get the answer.

Complete step by step answer:
According to the question, we have been given an arithmetic progression having the general term ${T_r}$ for $r = 1,2,3,....$.
We know that the ${n^{{\text{th}}}}$ term of an arithmetic progression is written as ${T_n} = a + \left( {n - 1} \right)d$, where $a$ is its first term and $d$ is its common difference. Based on this, we have:
$ \Rightarrow {T_r} = a + \left( {r - 1} \right)d{\text{ }}.....{\text{(1)}}$
Further, it is given that the ${m^{{\text{th}}}}$ term is $\dfrac{1}{n}$ i.e. ${T_m} = \dfrac{1}{n}$. Using formula for general term, we have:
$
   \Rightarrow {T_m} = a + \left( {m - 1} \right)d = \dfrac{1}{n} \\
   \Rightarrow a + \left( {m - 1} \right)d = \dfrac{1}{n}{\text{ }}.....{\text{(2)}} \\
 $
Also it is given that the ${n^{{\text{th}}}}$ term is $\dfrac{1}{m}$ i.e. ${T_n} = \dfrac{1}{m}$. Again using formula for general term, we have:
$
   \Rightarrow {T_n} = a + \left( {n - 1} \right)d = \dfrac{1}{m} \\
   \Rightarrow a + \left( {n - 1} \right)d = \dfrac{1}{m}{\text{ }}.....{\text{(3)}} \\
 $
Subtracting equation (3) from equation (2), we’ll get:
$
   \Rightarrow a + \left( {m - 1} \right)d - \left[ {a + \left( {n - 1} \right)d} \right] = \dfrac{1}{n} - \dfrac{1}{m} \\
   \Rightarrow a + md - d - a - nd + d = \dfrac{{m - n}}{{mn}} \\
   \Rightarrow \left( {m - n} \right)d = \dfrac{{m - n}}{{mn}} \\
 $
From both sides, $\left( {m - n} \right)$ will be cancelled, we’ll get:
$ \Rightarrow d = \dfrac{1}{{mn}}$
Putting the value $d = \dfrac{1}{{mn}}$ in equation (2), we’ll get:
$
   \Rightarrow a + \left( {m - 1} \right)\dfrac{1}{{mn}} = \dfrac{1}{n} \\
   \Rightarrow a + \dfrac{m}{{mn}} - \dfrac{1}{{mn}} = \dfrac{1}{n} \\
   \Rightarrow a + \dfrac{1}{n} - \dfrac{1}{{mn}} = \dfrac{1}{n} \\
   \Rightarrow a = \dfrac{1}{{mn}} \\
 $
Thus we have $a = \dfrac{1}{{mn}}$ and $d = \dfrac{1}{{mn}}$.
Now we have to determine ${T_{mn}}$. Putting $r = mn$ in equation, we have:
$ \Rightarrow {T_{mn}} = a + \left( {mn - 1} \right)d$
Putting $a = \dfrac{1}{{mn}}$ and $d = \dfrac{1}{{mn}}$, we’ll get:
$
   \Rightarrow {T_{mn}} = \dfrac{1}{{mn}} + \left( {mn - 1} \right)\dfrac{1}{{mn}} \\
   \Rightarrow {T_{mn}} = \dfrac{1}{{mn}} + \dfrac{{mn}}{{mn}} - \dfrac{1}{{mn}} \\
   \Rightarrow {T_{mn}} = 1 \\
 $
Thus the value of ${T_{mn}}$ is 1.

So, the correct answer is Option C.

Note: ${T_n} = a + \left( {n - 1} \right)d$ is the formula for the general term of an arithmetic progression. If an arithmetic progression is having $n$ terms with $a$ as its first term and $d$ is the common difference then the sum of its terms can be calculated using the formula:
$ \Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$