Question

If $m$th term of an AP is $\frac{1}{n}$ and $n$th is $\frac{1}{m}$ then show that $(mn{)}^{th}$ term of an AP is $1$.

Answer 1

We know,

$n^{th}$ term $t_n=a+(n-1)d$

$a=$ First term

$d=$ Common difference

$n=$ number of terms

$t_n=n^{th}$ term

Given,

$a_m=\frac{1}{n}\Rightarrow a+(m-1)d=\frac{1}{n}$.......(1)

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

(2)-(1)$\Rightarrow d=\frac{1}{mn}$

$\therefore a=\frac{1}{mn}$

Now, $mn$ th term,

$a+(mn-1)d=\frac{1}{mn}+1-\frac{1}{mn}=1$

Hence proved.