Question

If $ {{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}} $ and $ {{a}_{6}} $ are six arithmetic means between $ 3 \,and\, 31 $ , then $ {{a}_{6}}-{{a}_{5}} $ and $ {{a}_{1}}+{{a}_{6}} $ are respectively equals to:

• A. 5 and 34
• B. 4 and 35
• C. 4 and 34
• D. 4 and 36
• E. 6 and 36

Answer 1

Answer: (C)

The given AP is $ 3,{{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}},{{a}_{6}},31 $
$ \therefore $ $ 31=3+7d $
$ \Rightarrow $ $ d=4 $
$ \therefore $ $ {{a}_{1}}=3+4=7 $
$ {{a}_{5}}=a+5d=3+20=23 $ and $ {{a}_{6}}=a+6d=3+24=27 $
$ \therefore $ $ {{a}_{6}}-{{a}_{5}}=27-23=4 $ and $ {{a}_{1}}+{{a}_{6}}=7+27=34 $