Choosing Tickets in Arithmetic Progression

Let's consider the number of AP's which have the number $b$ as the middle number in the $AP$ .

If $b \leq 50$ , then we could have

• $1, b, 2b -1$
• $2, b, 2b-2$
• $\ldots$
• $b-1, b, b+1$

Thus, there are $b-1$ of them for each $b$ .
In total, there are $1 + 2 + 3 + \ldots + 50 = 1225$

Similarly, if $b \geq 51$ , then we could have

• $2b - 100, b, 100$
• $2b - 99, b, 99$
• $\ldots$
• $b-1, b, b+1$

Thus, there are $100 - b$ of them for each $b$ .
In total, there are $1 + 2 + 3 + \ldots + 50 = 1225$ .

Hence, there are $1225 + 1225 = 2450$ .

To solve this exercise, we need to sum all the possible triplets that forms an arithmetic progression of reason n between 1 and 100.

There are :

98 possiblities for n = 1

96 possibilities for n = 2

....

2 possibilities for n = 49 (above which there are no more solutions since 1 + 50 + 50 = 101)

Hence we must now calculate the sum :

S = 2 + 4 + 6 + 8 +....+ 96 + 98 S = 2 × 49 × 50 / 2 = 2450