Question

If m${}^{th}$ term of an H.P. is $n$ and n${}^{th}$ term is m then its mn${}^{th}$ term is

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{T_{m+n}}{T_m.T_n}$

Answer 1

Answer: (A)

$1$

Given that,$m^{th}$term of an H.P. is $n$ and $n^{th}$ term is $m$.

Therefore, $m^{th}$ term of an A.P. is $\frac{1}{n}$ and $n^{th}$ term is $\frac{1}{m}$.
Let, $a$ be the first term and $d$ be the common difference of the A.P

Now, $t_m=a+(m-1)d=\frac{1}{n}$ ....(1)

And $t_n=a+(n-1)d=\frac{1}{m}$ ....(2)

Subtracting (1) & (2), we get

$(m-n)d=\frac{1}{n}-\frac{1}{m}$

$\Rightarrow d=\frac{1}{mn}$
Substituting $d$ in (1), we get

$a+\frac{(m-1)}{mn}=\frac{1}{n}$ $\Rightarrow a=\frac{1}{mn}$
Therefore, $t_{mn}=a+(mn-1)d=\frac{1}{mn}+\frac{(mn-1)}{mn}=1$

and thus ${mn}^{th}$ term of H.P is $1$.