If mth term of an AP is 1/n and its nth term is 1/m , then show that its (mn)th term is 1

Options

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

Answer

D

Explanation

Let 'a' be the first term and 'd' be the common difference of an A.P
Now, Given `m^(th)` term `=(1)/(n)`
and `n^(th)" term "=(1)/(m)`
`impliesa+(m-1)d=(1)/(n)" and"`
`a+(n-1)d=(1)/(m)`
By subtracting the above two eqns. We get
`(m-1-n+1)d=(1)/(n)-(1)/(m)`
`implies(m-n)d=(m-n)/(mn)`
`impliesd=(1)/(mn)`
Now, `(mn)^(th)` term `=a+1-(1)/(mn)`
Now, `a=(1)/(m)-(n-1)d`
`=(1)/(m)-(n-1)(1)/(mn)=(1)/(m)-(n)/(mn)+(1)/(mn)=(1)/(mn)`
`:." "(mn)^(th)" term"=(1)/(mn)+1-(1)/(mn)=1`