A) 5, 9, 11, 13
B) 7, 11, 15, 19
C) 5, 11, 15, 22
D) 7, 15, 19, 21
Correct Answer: B
Solution :
Let four arithmetic means are \[{{A}_{1}},{{A}_{2}},\ {{A}_{3}}\]and \[{{A}_{4}}\]. So \[3,\ {{A}_{1}},\ {{A}_{2}},\ {{A}_{3}},\ {{A}_{4}},\ 23\] \[\Rightarrow \]\[{{T}_{6}}=23=a+5d\]\[\Rightarrow \]\[d=4\] Thus\[{{A}_{1}}=3+4=7,\ {{A}_{2}}=7+4=11,\ \] \[{{A}_{3}}=11+4=15,\,{{A}_{4}}=15+4=19\]
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