If p arithmetic means are inserted If $p$ arithmetic means $A_1$, $A_2$, ..., $A_p$, are inserted between $5$ and $41$ so that the relation is satisfied:
$$
\frac{A_3}{A_{(p-1)}}=\frac{2}{5}
$$
Find the value of $p/11$
I've tried using the formula that number of means $(n) = a+n[(b-a) /n+1]$ but my value of $p$ keeps coming $-12/13$ which is wrong. The correct ans that $p/11$ is 1
Pls help me out. 

When we insert p terms, the $n^{th}$ mean term is, 
$A_n = a + nd$
Here, 
$ d= \frac{last term -first term}{no. of terms - 1} = \frac{41 - 5}{p+1} = \frac{36}{p+1}$
$a = 5$,$a_n = 41$
$\frac{A_3}{A_{p-1}} = \frac{5 + 3\frac{36}{p+1}}{5 + (p-1)\frac{36}{p+1}} = \frac{2}{5}$
Taking LCM and solving, 
$\frac{5p+5 + 108}{5p + 5 + 36p - 36} = \frac{5p+113}{41p - 31}=\frac{2}{5}$
Cross multiplying,
$25p + 565 = 82p - 62$
$627 = 57p$
$\frac{627}{57} = 11 = p$
Or,$$ \frac{p}{11} = 1$$
