Question

There are $n$ Arithmetic means between $13$ and $61.$ If ratio of the first $A.M$. and $\left(n-1\right)$ the $A.M.$ may be $16:55,$ then find $n.$

Answer 1

Let $A_1,A_2,A_3,........A_n$ be the $n$ arithmetic means between $13$ and $61$.

Then, the common difference is given by:

$d=\frac{b-a}{n+1}$

$d=\frac{61-13}{n+1}$

$d=\frac{48}{n+1}$

Now,

$A_1=a+d=13+\frac{48}{n+1}$

$A_1=\frac{13n+61}{n+1}$

And

$A_n=a+nd=13+\frac{48n}{n+1}$

$A_n=\frac{61n+13}{n+1}$

Since,

$\frac{A_1}{A_n}=\frac{16}{55}$

$\frac{13n+61}{61n+13}=\frac{16}{55}$

$715n+3355=976n+208$

$261n=3147$

$n\approx 12$

Hence, this is the answer.