A) 310
B) 300
C) 320
D) None of these
Correct Answer: A
Solution :
\[{{n}^{th}}\] term of the series is \[20+(n-1)\,\left( -\frac{2}{3} \right)\]. For sum to be maximum, \[{{n}^{th}}\] term \[\ge 0\] \[\Rightarrow \]\[20+(n-1)\left( -\frac{2}{3} \right)\ge 0\]\[\Rightarrow \]\[n\le 31\] Thus the sum of 31 terms is maximum and is equal to \[\frac{31}{2}\left[ 40+30\times \left( -\frac{2}{3} \right) \right]=310\].
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