A) 5,9,11,13
B) 7,11,15,19
C) 5,11,15,22
D) 7,15,19,21
Correct Answer: B
Solution :
[b]: 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 \[{{A}_{1}}=3+4=7,{{A}_{2}}=7+4=11,\] \[{{A}_{3}}=11+4=15,{{A}_{4}}=15+4=19\]You need to login to perform this action.
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