question_answer

pth  term of the series\[\left( 3-\frac{1}{n} \right)+\left( 3-\frac{2}{n} \right)+\left( 3-\frac{3}{n} \right)+....\] will be

A) \[\left( 3+\frac{p}{n} \right)\]

B) \[\left( 3-\frac{p}{n} \right)\]

C) \[\left( 3+\frac{n}{p} \right)\]

D) \[\left( 3-\frac{n}{p} \right)\]

Correct Answer: B

Solution :

Given series \[\left( 3-\frac{1}{n} \right)+\left( 3-\frac{2}{n} \right)+\left( 3-\frac{3}{n} \right)+........\](A.P.) Therefore common difference \[d=\left( 3-\frac{2}{n} \right)-\left( 3-\frac{1}{n} \right)=-\frac{1}{n}\] and first term \[a=\left( 3-\frac{1}{n} \right)\] Now \[{{p}^{th}}\]term of the series \[=a+(p-1)d\] \[=\left( 3-\frac{1}{n} \right)+(p-1)\left( -\frac{1}{n} \right)=3-\frac{1}{n}+\frac{1}{n}-\frac{p}{n}=\left( 3-\frac{p}{n} \right)\]. Trick:  This question can also be done by inspection first\[-\frac{1}{n}\], second\[-\frac{2}{n}\], third\[-\frac{3}{n}\], therefore, \[{{p}^{th}}\]will be\[-\frac{p}{n}\]. Hence the result (3 is constant).