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If the m^{th} term of an A.P. is \frac{1}{n} and n^{th} term is \frac{1}{m} , then show that its  (mn)^{th} term is 1.

 

 

 

 
 
 
 
 

Answers (1)

Given that 

a_m=\frac{1}{n}, a_n=\frac{1}{m}

Let a and d be the first term and the common difference of the AP.

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

a+(n-1)d=\frac{1}{m} \;\;\;\;\;\; -(2)

Substracting the equation (1) and (2), we get 

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

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

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

If we put the value of d in eq (1),

a=\frac{1}{mn}

Let A be the  (mn)^{th}  term of the AP.

A=a+(mn-1)d

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

     =\frac{1}{mn} + 1 - \frac{1}{mn}

A=1

So, a_{mn} = 1

Hence proved.

 

 

Posted by

Ravindra Pindel

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