Between two numbers whose sum is $12$, an even number of A.M.s is inserted, the sum of these means exceeds their number by $20$. Find the number of means.

$$\begin{gathered} letthenumberofmeansintroduced=n \\ andfirsttermbeaandlastterm=b \\ a,A_1,A_2,..........,A_n,b \\ nowsumofmeans \\ =A_1+A_2+............+A_n \\ (\frac{n}{2})\cdot (a+b)=(\frac{n}{2})\cdot 12=6n \\ andasgiven6n=n+20 \\ \therefore 5n=20 \\ \therefore n=4 \end{gathered}$$