A) \[\frac{1}{6}\]
B) \[\frac{3}{7}\]
C) \[\frac{1}{7}\]
D) \[\frac{3}{5}\]
Correct Answer: B
Solution :
\[a+3d=\frac{5}{3}\]and \[a+7d=3\] Solving, \[a=\frac{2}{3},d=\frac{1}{3}\] 6th term of A.P. \[=a+5d=\frac{2}{3}+\frac{5}{3}=\frac{7}{3}\] Þ 6th term of H.P. =\[\frac{3}{7}\].
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