Problem Statement
If the $m^{\text{th}}$ term of an A.P. is $\frac{1}{n}$ and its $n^{\text{th}}$ term is $\frac{1}{m}$, show that its $(mn)^{\text{th}}$ term is $1$.
Verified Solution & Marking Scheme
Formulate Equations from Given Terms
Let the first term of the A.P. be $a$ and common difference be $d$.
- $T_m = a + (m - 1)d = \frac{1}{n}$ ...(1)
- $T_n = a + (n - 1)d = \frac{1}{m}$ ...(2)
Subtract to Find Common Difference d
Subtracting (2) from (1):
$(m - 1)d - (n - 1)d = \frac{1}{n} - \frac{1}{m}$
$(m - n)d = \frac{m - n}{mn}$
Since $m \neq n$, dividing by $(m - n)$:
$d = \frac{1}{mn}$
Substitute d to Find First Term a
From (1):
$a + (m - 1)\left(\frac{1}{mn}\right) = \frac{1}{n}$
$a = \frac{1}{n} - \frac{m - 1}{mn} = \frac{m - (m - 1)}{mn} = \frac{1}{mn}$
So $a = \frac{1}{mn}$ and $d = \frac{1}{mn}$.
Compute T_(mn)
$T_{mn} = a + (mn - 1)d = \frac{1}{mn} + (mn - 1)\left(\frac{1}{mn}\right)$
$= \frac{1 + mn - 1}{mn} = \frac{mn}{mn} = 1$
Hence proved.