Insert A.M.s between 7 and 71 in such a way that the 5th A.M. is 27.

Let $A_{1}, A_{2}, A_{3}, A_{4}, 27, A_{6} \ldots A_{n}$ be the $n$ arithmetic means between 7 and 71 . Thus, there are $(n+2)$ terms in all.

Thus, there are $(n+2)$ terms in all.

Let d be the common difference of the above A.P.

Now, a6 = 27

$\Rightarrow a+(6-1) d=27$

$\Rightarrow a+5 d=27$

$\Rightarrow d=4$

Also, $71=a_{n+2}$

$71=7+(n+2-1) 4$

$\Rightarrow 71=7+(n+1) 4$

$\Rightarrow n=15$

Therefore, there are 15 A.M.s.