Question

Out of $21$ tickets marked with numbers from $1$ to $21,$ three are drawn at random. Find the chance that the numbers on them are in A.P.

Answer 1

Total number of ways if common difference of the A.P. is to be $1,$ then the possible groups are $1,2,3;2,3,4;....19,20,21.$

If the common difference is $2$, then possible groups are $1,3,5;2,4,6;.....17,19,21$.

Proceeding in the same way, if the common difference is $10,$ then the possible group is $1,10,21.$

Thus, if the common difference of the A.P. is to be $11$, obviously there is no favourable case

Hence, total number of favourable cases $=19+17+15+...+3+1=100$

Hence, required probability $=\frac{100}{1330}=\frac{10}{133}$