Let [tex]\( T_r \)[/tex] be the [tex]\( r^{\text{th}} \)[/tex] term of an A.P. for [tex]\( r = 1, 2, \ldots \)[/tex].

Given that for some positive integers [tex]\( m \)[/tex] and [tex]\( n \)[/tex], we have [tex]\( T_m = \frac{1}{n} \)[/tex] and [tex]\( T_n = \frac{1}{m} \)[/tex],

Find [tex]\( T_{mn} \)[/tex].



Answer :

To solve for [tex]\( T_{mn} \)[/tex], the [tex]\( mn \)[/tex]-th term of the given arithmetic progression (AP) where [tex]\( T_r \)[/tex] denotes the [tex]\( r \)[/tex]-th term, given that:

[tex]\[ T_m = \frac{1}{n} \][/tex]
[tex]\[ T_n = \frac{1}{m} \][/tex]

we start by understanding the general form of the [tex]\( r \)[/tex]-th term of an AP:

[tex]\[ T_r = a + (r - 1)d \][/tex]

where [tex]\( a \)[/tex] is the first term and [tex]\( d \)[/tex] is the common difference.

Given [tex]\( T_m \)[/tex] and [tex]\( T_n \)[/tex], we write the following two equations using the general formula:

[tex]\[ T_m = a + (m - 1)d = \frac{1}{n} \][/tex]
[tex]\[ T_n = a + (n - 1)d = \frac{1}{m} \][/tex]

To find [tex]\( a \)[/tex] and [tex]\( d \)[/tex], we solve these two linear equations simultaneously.

Substitute [tex]\( T_m = \frac{1}{n} \)[/tex] and [tex]\( T_n = \frac{1}{m} \)[/tex] into the equations:

[tex]\[ a + (m - 1)d = \frac{1}{n} \][/tex]
[tex]\[ a + (n - 1)d = \frac{1}{m} \][/tex]

By eliminating [tex]\( a \)[/tex] from these equations, we subtract the second equation from the first:

[tex]\[ (a + (m - 1)d) - (a + (n - 1)d) = \frac{1}{n} - \frac{1}{m} \][/tex]

[tex]\[ (m - 1)d - (n - 1)d = \frac{1}{n} - \frac{1}{m} \][/tex]

Simplify the left side:

[tex]\[ (m - 1 - (n - 1))d = \frac{1}{n} - \frac{1}{m} \][/tex]

[tex]\[ (m - n)d = \frac{1}{n} - \frac{1}{m} \][/tex]

Notice the right side can be simplified using a common denominator:

[tex]\[ (m - n)d = \frac{m - n}{mn} \][/tex]

Thus, the [tex]\( m - n \)[/tex] terms cancel out:

[tex]\[ d = \frac{1}{mn} \][/tex]

Now substitute [tex]\( d \)[/tex] back into one of the original equations to find [tex]\( a \)[/tex]. Using the first equation:

[tex]\[ a + (m - 1)\frac{1}{mn} = \frac{1}{n} \][/tex]

[tex]\[ a + \frac{m - 1}{mn} = \frac{1}{n} \][/tex]

Clearing the fractions by multiplying through by [tex]\( mn \)[/tex]:

[tex]\[ a \cdot mn + (m - 1) = m \][/tex]

[tex]\[ a \cdot mn = m - (m - 1) \][/tex]

[tex]\[ a \cdot mn = 1 \][/tex]

[tex]\[ a = \frac{1}{mn} \][/tex]

With [tex]\( a \)[/tex] and [tex]\( d \)[/tex] determined, we can find [tex]\( T_{mn} \)[/tex]:

[tex]\[ T_{mn} = a + (mn - 1)d \][/tex]

Substitute [tex]\( a = \frac{1}{mn} \)[/tex] and [tex]\( d = \frac{1}{mn} \)[/tex]:

[tex]\[ T_{mn} = \frac{1}{mn} + (mn - 1)\frac{1}{mn} \][/tex]

Distribute and simplify:

[tex]\[ T_{mn} = \frac{1}{mn} + \frac{mn - 1}{mn} \][/tex]

Combine the fractions:

[tex]\[ T_{mn} = \frac{1 + (mn - 1)}{mn} \][/tex]

[tex]\[ T_{mn} = \frac{mn}{mn} \][/tex]

[tex]\[ T_{mn} = 1 \][/tex]

Thus, the term [tex]\( T_{mn} \)[/tex] is:

[tex]\[ T_{mn} = \frac{(mn - 1) + 1}{mn} = \frac{mn}{mn} = 1 \][/tex]