Sequence And Series Question 229
Question: pth term of the series $ ( 3-\frac{1}{n} )+( 3-\frac{2}{n} )+( 3-\frac{3}{n} )+…. $ will be
Options:
A) $ ( 3+\frac{p}{n} ) $
B) $ ( 3-\frac{p}{n} ) $
C) $ ( 3+\frac{n}{p} ) $
D) $ ( 3-\frac{n}{p} ) $
Show Answer
Answer:
Correct Answer: B
Solution:
Given series $ ( 3-\frac{1}{n} )+( 3-\frac{2}{n} )+( 3-\frac{3}{n} )+…….. $ (A.P.) Therefore common difference $ d=( 3-\frac{2}{n} )-( 3-\frac{1}{n} )=-\frac{1}{n} $ and first term $ a=( 3-\frac{1}{n} ) $ Now $ p^{th} $ term of the series $ =a+(p-1)d $ $ =( 3-\frac{1}{n} )+(p-1)( -\frac{1}{n} )=3-\frac{1}{n}+\frac{1}{n}-\frac{p}{n}=( 3-\frac{p}{n} ) $ . 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).
BETA
