The p^{th} term of the series left( 3 - frac{ | vedclass

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(B) The given series is $\left( 3 - \frac{1}{n} \right) + \left( 3 - \frac{2}{n} \right) + \left( 3 - \frac{3}{n} \right) + \dots$,which is an Arithme

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$\left( 3 - \frac{p}{n} \right)$

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(B) The given series is $\left( 3 - \frac{1}{n} \right) + \left( 3 - \frac{2}{n} \right) + \left( 3 - \frac{3}{n} \right) + \dots$,which is an Arithmetic Progression ($A$.$P$.).The first term $a = \left( 3 - \frac{1}{n} \right)$.The common difference $d = \left( 3 - \frac{2}{n} \right) - \left( 3 - \frac{1}{n} \right) = -\frac{2}{n} + \frac{1}{n} = -\frac{1}{n}$.The $p^{th}$ term of an $A$.$P$. is given by the formula $T_p = a + (p - 1)d$.Substituting the values:$T_p = \left( 3 - \frac{1}{n} \right) + (p - 1)\left( -\frac{1}{n} \right)$$T_p = 3 - \frac{1}{n} - \frac{p}{n} + \frac{1}{n}$$T_p = 3 - \frac{p}{n}$.Alternatively,by observation,the $k^{th}$ term is $\left( 3 - \frac{k}{n} \right)$. Therefore,the $p^{th}$ term is $\left( 3 - \frac{p}{n} \right)$.

The $p^{th}$ term of the series $\left( 3 - \frac{1}{n} \right) + \left( 3 - \frac{2}{n} \right) + \left( 3 - \frac{3}{n} \right) + \dots$ will be

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