A) 28
B) 31
C) 29.5
D) 26.5
Correct Answer: C
Solution :
\[{{a}_{1}}{{a}_{2}}{{a}_{3}}\] \[{{a}_{2}}=\frac{3a}{3}=13\] \[d=3\] sum of last four term = 178 \[{{a}_{n-3}}{{a}_{n-2}}{{a}_{n-1}}{{a}_{n}}\] Their mean \[=\frac{178}{4}=44.5\] \[{{a}_{n}}=44.5+1.5+3\] = 49 Median \[=\frac{10+49}{2}=\frac{59}{2}=29.5\]
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