A) \[\frac{3}{13}\]
B) \[\frac{6}{73}\]
C) \[\frac{3}{87}\]
D) \[\frac{6}{97}\]
Correct Answer: B
Solution :
(b): Terms in A.P. are: \[-\frac{1}{2},\frac{1}{6},\frac{5}{6},\frac{3}{2},........\] \[\therefore \]\[a=-\frac{1}{2};\] \[d=\frac{1}{6}-\left( -\frac{1}{2} \right)=\frac{1}{6}+\frac{1}{2}={}^{2}/{}_{3}\] \[\frac{1}{{{T}_{20}}}=-\frac{1}{2}+(20-1)\times \frac{2}{3}=-\frac{1}{2}+\frac{38}{3}=\frac{-3+76}{6}=\frac{73}{6}\] \[\therefore {{T}_{20}}=\frac{6}{73}\]
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