An even number of AM are inserted between two numbers whose sum is 13/6. If the sum of means exceeds their number by 1, what is the number of means?

Let the two numbers be a and b in which 2n A.M are inserted, then the A.P. will be

$a,A_1,A_2,\dots \dots ,A_{2n},b$

Here, first term = a, last term = b

$S=\frac{2n+2}{2}(a+b)\text{ and }a+b=\frac{13}{6}$

Total number of terms= 2n + 2 Sum of 2n + 2 terms

$$\begin{gathered} \mathrm{S}=\frac{2\mathrm{n}+2}{2}(\mathrm{a}+\mathrm{b}) \\ \mathrm{S}=(\mathrm{n}+1)(\mathrm{a}+\mathrm{b}) \end{gathered}$$

Now, sum of means ${\mathrm{A}}_1+{\mathrm{A}}_2+\dots .{\mathrm{A}}_{2n}=2\mathrm{n}+1$ (given)

And , Sum of means= sum of series  $-(a+b)$

$\begin{array}{l}\Rightarrow 2n+1=(n+1)(a+b)-(a+b)\\ \Rightarrow 2n+1=n(a+b)\end{array}$

$$\begin{gathered} \Rightarrow 2\mathrm{n}+1=\mathrm{n}\left(\frac{13}{6}\right) \\ \Rightarrow 13\mathrm{n}=12\mathrm{n}+6 \\ \Rightarrow \mathrm{n}=6 \end{gathered}$$

Number of means $=2\mathrm{n}=2\times 6=12$