Question

P^th term of the series $(3-\frac{1}{n})+(3-\frac{2}{n})+(3-\frac{3}{n})+........$ will be:

• A. $3+\frac{p}{n}$
• B. $3-\frac{p}{n}$
• C. $3+\frac{n}{p}$
• D. $3-\frac{n}{p}$

Answer 1

The correct option is B

$3-\frac{p}{n}$

Explanation for correct option

Calculating the pth term:

The given series is in arithmetic progression.

Here, first term $a_1=3-\frac{1}{n}$

Common difference:

$$\begin{gathered} d=a_2-a_1 \\ =3-\frac{2}{n}-3+\frac{1}{n} \\ =-\frac{1}{n} \end{gathered}$$

For the p^th term:

$$\begin{gathered} a_p=a+(p-1)d \\ =(3-\frac{1}{n})+(p-1)(-\frac{1}{n}) \\ =3-\frac{1}{n}-\frac{p}{n}+\frac{1}{n} \\ =3-\frac{p}{n} \end{gathered}$$

Therefore, the pth term is $3-\frac{n}{p}$.

Hence, the correct answer is option (B).