Sequence And Series Question 481
Question: The maximum sum of the series $ 20+19\frac{1}{3}+18\frac{2}{3}+……… $ is
Options:
A) 310
B) 300
C) 320
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ n^{th} $ term of the series is $ 20+(n-1),( -\frac{2}{3} ) $ . For sum to be maximum, $ n^{th} $ term $ \ge 0 $
$ \Rightarrow $ $ 20+(n-1)( -\frac{2}{3} )\ge 0 $
$ \Rightarrow $ $ n\le 31 $ Thus the sum of 31 terms is maximum and is equal to $ \frac{31}{2}[ 40+30\times ( -\frac{2}{3} ) ]=310 $ .
BETA
