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Byju's Answer

Standard XII

Mathematics

Arithmetic Progression

Let Tr be the...

Question

Let $T_{r}$ be the $r^{t h}$ term of an AP for $r = 1, 2, 3, \dots$ . If for some positive integers $m$ , $n$ we have $T_{m} = \frac{1}{n}$ and $T_{n} = \frac{1}{m}$ , then $T_{m n}$ equals

$\frac{1}{m n}$

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$\frac{1}{m} + \frac{1}{n}$

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$1$

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$0$

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Solution

The correct option is C
$1$

Explanation for the correct option:
Finding $T_{m n}$ :
Let the first term be ' $a$ ‘, common difference be ’ $d$ ' and $r^{t h}$ term is $T_{r}$ .
Let $m$ and $n$ be any positive integers then $m^{t h}$ term can be written as,
$\begin{array}{rcl} \Rightarrow & T_{m} = a + (m - 1) d \\ \Rightarrow & \frac{1}{n} = a + (m - 1) d \end{array}$
And $n^{t h}$ term can be written as,
$\begin{array}{rcl} \Rightarrow & T_{n} = a + (n - 1) d \\ \Rightarrow & \frac{1}{m} = a + (n - 1) d \end{array}$
By subtracting $n^{t h}$ term from $m^{t h}$ term on both sides, we get
$\begin{array}{rcl} \frac{1}{n} - \frac{1}{m} & = & a + (m - 1) d - (a - (n - 1) d) \\ \frac{\Rightarrow m - n}{m n} & = & a + (m - 1) d - a - (n - 1) d \\ \Rightarrow & \frac{m - n}{m n} = m d - d - n d + d \\ \Rightarrow & \frac{m - n}{m n} = m d - n d \\ \Rightarrow & \frac{m - n}{m n} = (m - n) d \\ \Rightarrow & \frac{1}{m n} = d \end{array}$
Thus, the value of $d = \frac{1}{m n}$ .
Now substitute the value of $d$ in one of the $n^{t h}$ term or $m^{t h}$ term of equation.
Substituting $d$ as $\frac{1}{m n}$ in the equation of $\begin{array}{rcl} \frac{1}{n} & = & a + (m - 1) d \end{array}$ .
$\begin{array}{rcl} \frac{1}{n} & = & a + (m - 1) \frac{1}{m n} \\ \Rightarrow & \frac{1}{n} = a + \frac{m}{m n} - \frac{1}{m n} \\ \Rightarrow & \frac{1}{n} = a + \frac{1}{n} - \frac{1}{m n} \\ \Rightarrow & 0 = a - \frac{1}{m n} \\ \Rightarrow & a = \frac{1}{m n} \end{array}$
The value of $a$ is obtained as $\frac{1}{m n}$ .
Finding the value of $m n^{t h}$ term,
$\Rightarrow T_{m n} = a + (m n - 1) d$
Substitute $a$ as $\frac{1}{m n}$ and $d$ as $\frac{1}{m n}$ in the term $T_{m n} = a + (m n - 1) d$
$\begin{array}{rcl} T_{m n} & = & \frac{1}{m n} + (m n - 1) \frac{1}{m n} \\ = & \frac{1}{m n} + \frac{m n}{m n} - \frac{1}{m n} \\ = & \frac{m n}{m n} \\ = & 1 \end{array}$
Therefore, the value of $T_{m n}$ is equal to $1$ .
Hence, the correct answer is option (C).

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