If the $m^{\text{th}}$ term of an arithmetic progression is $\frac{1}{n}$ and the $n^{\text{th}}$ term is $\frac{1}{m}$, then the $mn^{\text{th}}$ term of this progression will be

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D - $1$. Subtracting the mth term from the nth term gives d = 1/(mn), and then a = 1/(mn). The mn-th term = a + (mn - 1)d = mn/(mn) = 1.

Answer

A - $\frac{1}{mn}$

Answer

B - $\frac{m}{n}$

Answer

C - $\frac{n}{m}$

If the \(m^{\text{th}}\) term of an arithmetic progression is \(\frac{1}{n}\) and the \(n^{\text{th}}\) term is \(\frac{1}{m}\), then the \(mn^{\text{th}}\) term of this progression will be

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