If a1,a2,a3,a4,a5 and a6 are six arithmetic means between 3 and 31 , then a6-a5 and a1+a6 are respectively equals to:

Question

If a1,a2,a3,a4,a5 and a6 are six arithmetic means between 3 and 31 , then a6-a5 and a1+a6 are respectively equals to: (A) 5 and 34 (B) 4 and 35 (C) 4 and 34 (D) 4 and 36

Tardigrade Admin · Accepted Answer

The given AP is 3,a1,a2,a3,a4,a5,a6,31 ∴ 31=3+7d ⇒ d=4 ∴ a1=3+4=7 a5=a+5d=3+20=23 and a6=a+6d=3+24=27 ∴ a6-a5=27-23=4 and a1+a6=7+27=34

Q. If $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ and $a_{6}$ are six arithmetic means between $3 an d 31$ , then $a_{6} - a_{5}$ and $a_{1} + a_{6}$ are respectively equals to:

5 and 34

4 and 35

4 and 34

4 and 36

6 and 36

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

Q. If a1​,a2​,a3​,a4​,a5​ and a6​ are six arithmetic means between 3and31 , then a6​−a5​ and a1​+a6​ are respectively equals to: