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

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

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

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(B) The given series is $\left( 3 - \frac{1}{n} \right) + \left( 3 - \frac{2}{n} \right) + \left( 3 - \frac{3}{n} \right) + \dots$ This is an Arithmetic Progression ($A$.$P$.). Here,the first term $a = \left( 3 - \frac{1}{n} \right)$. The common difference $d = \left( 3 - \frac{2}{n} \right) - \left( 3 - \frac{1}{n} \right) = -\frac{1}{n}$. The $p^{th}$ term of an $A$.$P$. is given by $T_p = a + (p - 1)d$. Substituting the values,we get: $T_p = \left( 3 - \frac{1}{n} \right) + (p - 1)\left( -\frac{1}{n} \right)$ $T_p = 3 - \frac{1}{n} - \frac{p}{n} + \frac{1}{n}$ $T_p = 3 - \frac{p}{n}$. Alternatively,by inspection: $1^{st}$ term: $3 - \frac{1}{n}$ $2^{nd}$ term: $3 - \frac{2}{n}$ $3^{rd}$ term: $3 - \frac{3}{n}$ Therefore,the $p^{th}$ term is $3 - \frac{p}{n}$.

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

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