Let $${T_r}$$ be the $${r^{th}}$$ term of an A.P. whose first term is $$a$$ and common difference is $$d$$. If for some positive integers $$m,n,\,\,m \ne n,\,{T_m} = \frac{1}{n}{\text{ and }}{T_n} = \frac{1}{m},$$       then $$a - d$$  equals

Solution :
$$\eqalign{ & {T_m} = a + \left( {m - 1} \right)d = \frac{1}{n}\,\,\,......\left( 1 \right) \cr & {T_n} = a + \left( {n - 1} \right)d = \frac{1}{m}\,\,\,......\left( 2 \right) \cr & \left( 1 \right) - \left( 2 \right)\,\, \Rightarrow \left( {m - n} \right)d = \frac{1}{n} - \frac{1}{m} \cr & \Rightarrow \,d = \frac{1}{{mn}} \cr & {\text{From }}\left( 1 \right)\,\,a = \frac{1}{{mn}} \cr & \Rightarrow \,a - d = 0 \cr} $$