If p arithmetic means are inserted

Question

If $p$ arithmetic means $A_1$, $A_2$, ..., $A_p$, are inserted between $5$ and $41$ so that the relation below is satisfied-:

$$ \frac{A_3}{A_{(p-1)}}=\frac{2}{5} $$

Find the value of $p/11$.

I have attempted using the formula for the number of means, that is, $(n) = a + n[(b-a) /n+1]$. However, my value of $p$ keeps coming out as $-12/13$, which is wrong.

The correct answer is that $p/11$ is 1. Please help me out.

Answer 1

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,$$ \frac{p}{11} = 1$$