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In an AP, the $m^{\text {th }}$ term $1 / n$ and $n^{\text {th }}$ term is $1 / m$. What is its $(m n)^{\text {th }}$ term?
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Let 'a' be the first term and 'd' be the common difference of an A.P
Now, Given $\mathrm{m}^{\text {th }}$ term $=\frac{1}{n}$
and $\mathrm{n}^{\text {th }}$ term $=\frac{1}{m}$
$\Rightarrow \mathrm{a}+(\mathrm{m}-1) \mathrm{d}=\frac{1}{n}$ and
$a+(n-1) d=\frac{1}{m}$
By subtracting the above two eqns, we get
$(\mathrm{m}-1-\mathrm{n}+1) \mathrm{d}=\frac{1}{n}-\frac{1}{m}$
$\Rightarrow(\mathrm{m}-\mathrm{n}) \mathrm{d}=\frac{m-n}{m n}$
$\Rightarrow d=\frac{1}{m n}$
Now, $(\mathrm{mn})^{\text {th }}$ term $=\mathrm{a}+(\mathrm{mn}-1) \mathrm{d}$
$=a+(m n-1) \frac{1}{m n}=a+1-\frac{1}{m n}$
Now, $a=\frac{1}{m}-(n-1) d$
$=\frac{1}{m}-(n-1) \frac{1}{m n}=\frac{1}{m}-\frac{n}{m n}+\frac{1}{m n}=\frac{1}{m n}$
$\therefore(\mathrm{mn})^{\text {th }}$ term $=\frac{1}{m n}+1-\frac{1}{m n}=1$
Now, Given $\mathrm{m}^{\text {th }}$ term $=\frac{1}{n}$
and $\mathrm{n}^{\text {th }}$ term $=\frac{1}{m}$
$\Rightarrow \mathrm{a}+(\mathrm{m}-1) \mathrm{d}=\frac{1}{n}$ and
$a+(n-1) d=\frac{1}{m}$
By subtracting the above two eqns, we get
$(\mathrm{m}-1-\mathrm{n}+1) \mathrm{d}=\frac{1}{n}-\frac{1}{m}$
$\Rightarrow(\mathrm{m}-\mathrm{n}) \mathrm{d}=\frac{m-n}{m n}$
$\Rightarrow d=\frac{1}{m n}$
Now, $(\mathrm{mn})^{\text {th }}$ term $=\mathrm{a}+(\mathrm{mn}-1) \mathrm{d}$
$=a+(m n-1) \frac{1}{m n}=a+1-\frac{1}{m n}$
Now, $a=\frac{1}{m}-(n-1) d$
$=\frac{1}{m}-(n-1) \frac{1}{m n}=\frac{1}{m}-\frac{n}{m n}+\frac{1}{m n}=\frac{1}{m n}$
$\therefore(\mathrm{mn})^{\text {th }}$ term $=\frac{1}{m n}+1-\frac{1}{m n}=1$
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