Arithmetic progression 3.

Let $A_1,A_2,...,A_{2n}$ be $2n$ arithmetic means between $a$ and $b$ .

$\Rightarrow A_1+A_2+...+A_{2n}=2n×\left(\dfrac{a+b}{2}\right)$

$A_1+A_2+...+A_{2n}=n(a+b)=\dfrac{13}{6}n$

$2n+1=\dfrac{13}{6}n$ $[ \since A_1+A_2+...+A_{2n}=2n+1(\text{given})]$

$\therefore n=6$

Let the two numbers be $a$ (first term) and $l$ (second term).
$a + l = \dfrac{13}{6}\\ 6a + 6l = 13 \longrightarrow \boxed{1}$

Since an "even" number of arithmetic means are inserted, let the number of AMs inserted be $2n$ . Now, their sum exceeds their number by $1$ . The arithmetic means also follow arithmetic progression which is $A_1 + A_2 + A_3 + \cdots + A_{2n} = \dfrac{2n}{2} × \left(A_{2n} + A_1\right)$ . Now sum of terms equidistant from the back and front of an AP is equal. Here $a , A_1 , A_2 , A_3 , \cdots , A_{2n} , l$ form an arithmetic progression. Here $a$ is first term and $l$ is the last term. While, $A_1$ is the second term and $A_{2n}$ is the second last term. So, $A_1 + A_{2n} = a + l$ .

Inserting the above result in the sum of the arithmetic means we get:-
$S_{AM} = \dfrac{2n}{2} × \left(a + l\right)= n\left(a + l\right)$ .

Substituting $\boxed{1}$ in the above equation we get:-
$n\left(\dfrac{13}{6}\right) = \dfrac{13n}{6}$ .

Now, this value is one more than the number of arithmetic means inserted.
So, $\dfrac{13n}{6} = 2n + 1\\ 13n = 12n + 6\\ n = \color{#69047E}{\boxed{6}}$