Insert $20$ $AM$ between $2$ and $86$. Then find first mean.

Hint: All the terms placed between $2$ and $86$ are in $A.P$. So, first find common differences then find the series & then first mean.

Complete step-by-step answer:
In between $2$ and $86$, $20$ terms are to be inserted. Total number of terms is $22$.
From $A.P$ we know ${n^{th}}$ term $T_n$,
$T_n = a + (n - 1)d$
From the question, we take $T_n$ = $86$, First term $(a) = 2$ and $n = 22$.
$\therefore {T_{22}} = a + (n - 1)d$
$\Rightarrow 86 = 2 + (22 - 1)d$
$\Rightarrow 84 = 21d$
$\therefore d = \dfrac{{84}}{{21}} = 4$
The numbers which are in $A.P$, they were only differ by their common difference.
So, first mean we will put $ = a + 4 = 2 + 4 = 6$
$\therefore $ First mean $ = 6$
So, the required series is 2, 6, 10, 14,......... 86.
$\therefore $ The required mean is 6, 10, 10, 14, 18,....... 82.

Note: After inserting $20$ means all the numbers have the same common difference. So, after finding common differences we are able to write a series and therefore the first input between $2$ and $86$ is equal to the required mean.