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

Question

Dr. Abubacker · Accepted Answer

(c)
Given, AP is $3,\ a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},31$
$	herefore\ 31 = 3 + 7d$
$\Rightarrow d = 4$
$	herefore\ a_{1} = 3 + 4 = 7$
$a_{5} = a + 5d = 3 + 20 = 23$
and $a_{6} = a + 6d = 3 + 24 = 27$
$	herefore\ a_{6} - a_{5} = 27 - 23 = 4$
and $a_{1} + a_{6} = 7 + 27 = 34$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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