$\text{ There are }\mathrm{n}\text{-A.M is between }7\text{ and }85{\text{ such that (n-3) }}^{\text{th }}\text{ mean : n }{\mathrm{th}}_{}\mathrm{mean} \mathrm{is}as$$\text{11:24 then ‘n’ is}$

$7,{\mathrm{A}}_1,{\mathrm{A}}_2,\dots {\mathrm{A}}_{\mathrm{n}},85\text{ are in A.P }$

$\begin{array}{l}85=\left(n+2{)}^{th}\text{ term }=a+(n+1)d\right.\\ (n+1)d=78\\ d=\frac{78}{n+1}\end{array}$

$\begin{array}{l}\text{ Given that }\frac{(n-3{)}^{th}A\cdot M}{n^{th}A\cdot M}=\frac{11}{24}\Rightarrow \frac{7+(n-3)d}{7+(n+1-1)d}=\frac{11}{24}\\ \Rightarrow 168+24(n-3)d=77+11nd\\ \Rightarrow 91=d\left(11n-24n+72\right)\\ \Rightarrow 91\left(n+1\right)=78\left(72-13n\right)\\ \Rightarrow 1105n=5525\\ \Rightarrow n=5\end{array}$