Probability of selecting three even number in A.P.(Arithmetic Progression) among first 100 Natural number From first 100 natural numbers, 3 numbers are selected. If these three numbers are in A.P., then find the probability that these numbers are even.
My approach is as follow, selection of three number from first 100 natural numbers is $^{100}C_3$ which is equal to 161700.
Let the third term be represented by the series $a+2d=T$ where T is less than 100, where a and d are Natural Numbers
$a+2d \le100$
As the numbers are even a=2c and d=2e
$c+2e \le50$
$c\le50-2e$
e has values from 1 to 24
Total number of cases are $2*(1+2+3+..+24)=600$
Series can be reversed also hence number of cases are 1200, but the answer is $\frac{1}{66}$, which means that the number of cases are 1225 where the five cases are missing
 A: "From first $100$ natural numbers, $3$ numbers are selected. If these three numbers are in A.P., then find the probability that these numbers are even."
According to this text a conditional probability is asked for. Therefore we do not have to count the possible triple selections from $[100]$, but only the possible A.P. triple selections from $[100]$, and to analyze how many of these are all-even triples.
We obtain an A.P. triple by choosing either two even numbers and their average or two odd numbers and their average. Therefore we obtain $2\cdot{50\choose2}=2450$ possible A.P. triples, all of them equally probable. 
An all-even A.P. triple can be halved termwise, and in this way becomes an A.P. triple from $[50]$. There are $2\cdot{25\choose2}=600$ such triples.
The probability $p$ you are after therefore is given by
$$p={600\over2450}={12\over49}\ .$$
A: We can pick  a 3 number sequence in $^{100}C_3$ ways. Now, if the numbers are in AP and all are even, the first and the last terms of the AP have to be even. We have 50 even numbers in the first 100 natural numbers and we can pick 2 in $^{50}C_2$ ways. Therefore, the probability is:
$$P = \frac{2 \cdot \space ^{50}C_2}{^{100}C_3} = \frac 1 {66}$$
