Question

Let ${T_r}$ be the rth term of an A.P. whose first term is an $a$ and common difference is $d$. if for some positive integers m, n such that m is not equal to $n$, \[{T_m} = \dfrac{1}{n}\] and \[{T_n} = \dfrac{1}{m}\] then a-d equals to
A. 1
B. 0
C. \[\dfrac{1}{{mn}}\]
D. \[\dfrac{1}{n} + \dfrac{1}{m}\]

Answer 1

Hint: We are given the rth term of an A.P. also the first term and the common difference is given to us. We will use all the given data and will try to express the value of and $d$ in the form of $m$ and $n$. Then we will use the formula used in an A.P. for rth term. Then once we get the value of a and d we can easily find the difference.

Complete step by step answer:
Given that, a is the first term and d is the common difference. We know that,
\[{T_n} = a + \left( {n - 1} \right)d\] and similarly \[{T_m} = a + \left( {m - 1} \right)d\]
But it is given that, \[{T_m} = \dfrac{1}{n}\] and \[{T_n} = \dfrac{1}{m}\]
So on equating we get,
\[{T_m} = a + \left( {m - 1} \right)d = \dfrac{1}{n} \\
\Rightarrow {T_n} = a + \left( {n - 1} \right)d = \dfrac{1}{m} \\ \]
Now we will perform subtraction as,
\[ \Rightarrow \dfrac{1}{n} - \dfrac{1}{m} = a + \left( {m - 1} \right)d - \left[ {a + \left( {n - 1} \right)d} \right]\]
Now opening the brackets on RHS and taking LCM on LHS we get,
\[ \Rightarrow \dfrac{{m - n}}{{mn}} = a + \left( {m - 1} \right)d - a - \left( {n - 1} \right)d\]

On cancelling a from RHS,
\[ \Rightarrow \dfrac{{m - n}}{{mn}} = \left( {m - 1} \right)d - \left( {n - 1} \right)d\]
Taking d common,
\[ \Rightarrow \dfrac{{m - n}}{{mn}} = d\left( {m - 1 - n + 1} \right)\]
Cancelling 1,
\[ \Rightarrow \dfrac{{m - n}}{{mn}} = d\left( {m - n} \right)\]
Cancelling the common bracket,
\[ \Rightarrow \dfrac{1}{{mn}} = d\]
So this is the value of d.
Now we will put the value of d in the equation of \[{T_m}\]
\[a + \left( {m - 1} \right)\dfrac{1}{{mn}} = \dfrac{1}{n}\]

Separating the terms,
\[a + \dfrac{m}{{mn}} - \dfrac{1}{{mn}} = \dfrac{1}{n}\]
On solving the fractions,
\[a + \dfrac{1}{n} - \dfrac{1}{{mn}} = \dfrac{1}{n}\]
Cancelling the common terms,
\[a = \dfrac{1}{{mn}}\]
This is the value of a.
As we can see that the value of a and d is same then,
\[a - d = \dfrac{1}{{mn}} - \dfrac{1}{{mn}} = 0\]

So , option B is the correct answer.

Note:Here the most important thing one should know is the formula for rth term of an A.P. only. Because we have no other way to proceed. Sometimes we can reach the proof with the help of options. But not in this case. Also a single mistake while writing the suffixes $m$ and $n$ will cause the wrong solution.