Question

Please explain this problem step by step.
Will mark as brainliest if I'm satisfied with the answer.
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Answer 1

Answer:

solution mein koi doubt aaye to puch lena

Answer 2

Since $\displaystyle\sf{m^{th}}$ term is $\displaystyle\sf{\dfrac{1}{n},}$ we have,

$\displaystyle\longrightarrow\sf{a+(m-1)d=\dfrac{1}{n}\quad\quad\dots(i)}$

And since $\displaystyle\sf{n^{th}}$ term is $\displaystyle\sf{\dfrac{1}{m},}$ we have,

$\displaystyle\longrightarrow\sf{a+(n-1)d=\dfrac{1}{m}\quad\quad\dots(ii)}$

Now subtract (ii) from (i).

$\displaystyle\longrightarrow\sf{(i)-(ii)}$

$\displaystyle\longrightarrow\sf{[a+(m-1)d]-[a+(n-1)d]=\dfrac{1}{n}-\dfrac{1}{m}}$

$\displaystyle\longrightarrow\sf{a+(m-1)d-a-(n-1)d=\dfrac{m-n}{mn}}$

$\displaystyle\longrightarrow\sf{[(m-1)-(n-1)]d=\dfrac{m-n}{mn}}$

$\displaystyle\longrightarrow\sf{(m-1-n+1)d=\dfrac{m-n}{mn}}$

$\displaystyle\longrightarrow\sf{(m-n)d=\dfrac{m-n}{mn}}$

$\displaystyle\longrightarrow\sf{d=\dfrac{1}{mn}}$

Now we put $\displaystyle\sf{d=\dfrac{1}{mn}}$ in (i), i.e.,

$\displaystyle\longrightarrow\sf{a+(m-1)\cdot\dfrac{1}{mn}=\dfrac{1}{n}}$

$\displaystyle\longrightarrow\sf{a+\dfrac{m-1}{mn}=\dfrac{1}{n}}$

$\displaystyle\longrightarrow\sf{a=\dfrac{1}{n}-\dfrac{m-1}{mn}}$

$\displaystyle\longrightarrow\sf{a=\dfrac{m}{mn}-\dfrac{m-1}{mn}}$

$\displaystyle\longrightarrow\sf{a=\dfrac{m-(m-1)}{mn}}$

$\displaystyle\longrightarrow\sf{a=\dfrac{m-m+1}{mn}}$

$\displaystyle\longrightarrow\sf{a=\dfrac{1}{mn}}$

Now we have $\displaystyle\sf{a=d=\dfrac{1}{mn}.}$

Now, let's find $\displaystyle\sf{(mn)^{th}}$ term.

$\displaystyle\longrightarrow\sf{a_{mn}=a+(mn-1)d}$

$\displaystyle\longrightarrow\sf{a_{mn}=\dfrac{1}{mn}+(mn-1)\cdot\dfrac{1}{mn}}$

$\displaystyle\longrightarrow\sf{a_{mn}=\dfrac{1}{mn}+\dfrac{mn-1}{mn}}$

$\displaystyle\longrightarrow\sf{a_{mn}=\dfrac{1+mn-1}{mn}}$

$\displaystyle\longrightarrow\sf{a_{mn}=\dfrac{mn}{mn}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{a_{mn}=1}}}$

Hence Proved!