Question

If $P$ arithmetic means are inserted between $5$ and $41$ such that $\frac{A_3}{A_{P-1}}=\frac{2}{5}$, where $A_P$ represents the $P^{th}$ mean, then the value of $P$ is

Answer 1

$P$ A.M.'s are inserted between $5$ and $41$, so
$d=\frac{41-5}{P+1}=\frac{36}{P+1}$
Now,
$\frac{A_3}{A_{P-1}}=\frac{2}{5}\Rightarrow \frac{5+3d}{41-2d}=\frac{2}{5}\Rightarrow 25+15d=82-4d\Rightarrow d=3\Rightarrow \frac{36}{P+1}=3\therefore P=11$