Let `T` be the `r`th term of an A.P. whose first term is `a` and conmon difference is `d`. If for some positive integers `m ,n,` `T_(n)= (1)/(m) , T_(m) = (1)/(n) ` then `(a – d)` equals

Options

A. `(1)/(mn)`
B. `(1)/(m)+(1)/(n)`
C. 1
D. 0

Answer

C

Explanation

Let a be the term and d be the common difference of the given AP. Then,
`T_(m)=(1)/(n)rArra=(m+1)d=(1)/(n)` . . . (i)
`and,T_(n)=(1)/(m)rArra+(n-1)d=(1)/(m)` . . . (ii)
Solving (i) and (ii), we get `d=(1)/(mn)anda=(1)/(mn)`
`:." "T_(mn)=a+(mn-1)d=(1)/(mn-1)(1)/(mn)=1`