The sum of two numbers is $2\frac{1}{6}.$ An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Find the number of means inserted.

Let a and b be the two numbers so that $a+b\frac{13}{6},$ given $\left(1\right)$
Let 2n $\left(even\right)$ means be inserted between them whose sum $=\left(2n+1\right)$ given $...\left(2\right)$
Now we know that in an A.P., sum of 2n means $=2n\times \left(singleA.M,\right)$ or $2n+1=2n\left(\frac{a+b}{2}\right)$ by $\left(2\right)$ or $2n+1=n.\frac{13}{6},$ by $\left(1\right)$ or $12n+6=13n,\therefore n=6.$
Hence the number of means inserted $=2n=12.$