Out of 21 tickets consecutively numbered, there are drawn at random. Find the probability that the numbers on them are in AP is a / b, then (14a - b) is ......... .

Any three ticket8 out of21 tickets can be chosen is ${ }^{21}{\mathrm{C}}_3$ ways
For the favorable choice, if the chosen numbers are a, b and c, a < b < c then we should have $\frac{\mathrm{a}+\mathrm{c}}{2}=\mathrm{b}.$
Obviously either both o and care even or both are odd and then 6 is fixed 
Hence, for the favorable choice, we have to choose two numbers from 1 to 21, which are either both even or both odd.
This can be done in ${ }^{11}{\mathrm{C}}_2{+}^{10}{\mathrm{C}}_2$, ways. 

Hence, required probability $=\frac{{ }^{11}{\mathrm{C}}_2{+}^{10}{\mathrm{C}}_2}{{ }^{21}{\mathrm{C}}_3}=\frac{10}{133}$

$\Rightarrow \frac{\mathrm{a}}{\mathrm{b}}\Rightarrow 14\mathrm{a}-\mathrm{b}=140-133=7$