If the mth term of an AP is 1/n and the nth term is 1/m, then

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Q) If the mth term of an AP is 1/n and the nth term is 1/m, then show that the sum of (mn)th term is $\frac1 2$ (mn+1).

Ans:

Let first term of the given AP = a

Common difference of the given AP = d

Then,

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

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

Subtracting equation (ii) from equation (ii),we get

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

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

$=md-\cancel d-nd+\cancel d=\frac{1}{n}-\frac{1}{m}$

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

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

$\Rightarrow \cancel{(m-n)}d=\frac{\cancel{m-n}}{mn}$

$\Rightarrow d=\frac{1}{mn}$

Putting $d=\frac{1}{mn}$ in equation (i), we get

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

$\Rightarrow a+\cancel m\cdot\frac{1}{\cancel {{m}}{n}}- \frac{1}{mn} =\frac{1}{n}$

$\Rightarrow a+\cancel{\frac{1}{n}}- \frac{1}{mn} =\cancel{\frac{1}{n}}$

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

Next, we know that the sum of n terms of an AP is given by: $S_n = \frac{n}{2} (2a + (n - 1) d)$

Therefore the sum of mn terms will be given by: $S_{mn} = \frac{mn}{2} (2a + (mn - 1) d)$

By substituting values of a and d in the above formula, we get:

$S_{mn} = \frac{mn}{2} (2(\frac{1}{mn})+ (mn - 1) (\frac{1}{mn}))$

$S_{mn} = \frac{mn}{2}( \frac{2}{mn}+\cancel{mn}\cdot\frac{1}{\cancel{mn}} - \frac{1}{mn})$

$S_{mn} = \frac{mn}{2}( \frac{1}{mn}+1)$

$S_{mn} = \frac{1}{2}( \frac{\cancel{mn}}{\cancel{mn}}+mn)$

$S_{mn} = \frac{1}{2}(mn+1)$

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