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In an A.P., the n^th term is $\frac{1}{m}$ and the m^th term is $\frac{1}{n}.$ Find (i) $(mn)^{th}$ term, (ii) sum of first $(mn)$ terms.

Answers (1)

Given $\rightarrow a_{n}=\frac{1}{m}$

$a_{m}=\frac{1}{n}$

(i) To find $\Rightarrow a_{mn}=?$

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

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

Substract (ii) from (i)

$\frac{1}{m}-\frac{1}{n}=\left [ (n-1)-(m-1) \right ]d$

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

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

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

Putting value of 'd ' in eq (i)

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

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

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

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

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

$\Rightarrow a_{mn}=1$

(ii) To find $S_{mn}\; ?$

$S_{mn}= \frac{mn}{2}\left [ a_{mn}+a \right ]$

$=\frac{mn}{2}\left [ 1+\frac{1}{mn} \right ]$

$=\frac{mn}{2}+\frac{1}{2}$

$S_{mn}\Rightarrow \frac{mn+1}{2}$

Posted by

Safeer PP

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