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

From the series, $d=\frac{-1}{n},a=3-\frac{1}{n}$
$\therefore a_p=a+(p-1)d$

$\Rightarrow 3-\frac{1}{n}+(p-1)\frac{-1}{n}$
$\Rightarrow 3-\frac{1}{n}-\frac{p}{n}+\frac{1}{n}$

$\therefore p^{th}$ term $=3-\frac{p}{n}$