Sequence And Series Question 396
Question: The maximum sum of the series $ 20+19\frac{1}{3} $ $ +18\frac{2}{3}+18+…. $ is
Options:
A) 300
B) 310
C) $ 311\frac{2}{3} $
D) $ 333\frac{1}{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The given series is arithmetic whose first term = 20, and common difference $ =-\frac{2}{3} $ As the common difference is negative the terms will become negative after some stage. So the sum is maximum when all positive terms are added Now, for the positive terms $ x_{n}\ge 0\Rightarrow 20+(n-1)\times -\frac{2}{3}\ge 0 $
$ \Rightarrow ,60-2(n-1)\ge 0\Rightarrow n\le 31. $
$ \therefore $ The first 31 terms are non- negative
$ \therefore $ Maximum sum $ =S_{31}=\frac{31}{2}[ 2\times 20+(31-1)\times -\frac{2}{3} ]=310 $
BETA
