If $a,a_1,a_2,\dots ,a_{10}, b$ are in AP and $a,h_1,h_2,\dots \dots ,h_{10},b\text{ are in HP such that }{a}_1+a_2+\dots +a_{10}=25$ and $\frac{1}{{\mathrm{h}}_1}+\frac{1}{{\mathrm{h}}_2}+\dots +\frac{1}{{\mathrm{h}}_{10}}=\frac{25}{6}$, 

then a and b are

We know that the sum of n arithmetic means between two numbers is equal to n times their single mean.

$\begin{array}{l}\therefore a_1+a_2+\dots +a_{10}=10\left(\frac{a+b}{2}\right)\\ 25=5(a+b)\therefore a+b=5\\ a,h_1,h_2,\dots ,h_{10},b\text{ are in HP }\\ \Rightarrow \text{ So }1/a,1/h_1\dots 1/h_{10},1/b\text{ are in AP }\\ \therefore \frac{1}{h_1}+\frac{1}{h_2}+\dots +\frac{1}{h_{10}}=10\left(\frac{\frac{1}{a}+\frac{1}{b}}{2}\right)\\ \frac{25}{6}=5\times \frac{a+b}{ab}\Rightarrow ab=6\end{array}$

So, a and bare the roots of the equation $t^2-5t+6=0$

So, a and b are 2 and 3 .