If the m^{th} term of an A.P. be `1/n` and n^{th} term be `1/m`, then show that its (mn)^{th} term is 1.

#### Solution 1

Let a and d be the first term and common difference respectively of the given A.P.

Then,

1/n= m^{th} term ⇒ 1/n = a + (m – 1) d ….(i)

1/m = n^{th} term ⇒ 1/m = a + (n – 1) d ….(ii)

On subtracting equation (ii) from equation (i), we get

`\frac{1}{n}-\frac{1}{m}=( mn)d`

`\Rightarrow \frac{m-n}{mn}=( mn)d\Rightarrow d=\frac{1}{mn}`

`\frac{1}{n}=a+\frac{(m-1)}{mn}\Rightarrow a=\frac{1}{mn}`

∴ (mn)^{th} term = a + (mn – 1) d

`=\frac{1}{mn}+(mn-1)\frac{1}{mn}=1`

#### Solution 2

Let a and d be the first term and common difference respectively of the AP, respectively.

Then,

mth term =1/n

`=> a + (m - 1)d = 1/n`

`a = 1/n - (m -1)d` .....(1)

`"nth term"= 1/m`

`=> a + (n - 1)d = 1/m`

`=> 1/n -(m - 1)d + (n - 1)d = 1/m` [From 1]

`=> d(n - 1 - m + 1) = 1/m - 1/n`

`=> d(n - m) = (n - m)/mn`

`=> d = 1/"mn"`

Putting d = 1/mn in (1) we have

`a = 1/m - (m - 1) 1/(mn)`

`= 1/n - m/(mn) + 1/(mn)`

`= 1/n - 1/n + 1/(mn) = 1/(mn)`

`:. t_"mn" = a + (mn - 1)d = 1/(mn) + (mn -1) 1/(mn) = 1`