The question asks how many sets of three-digit numbers exist that are divisible by 5 and form an arithmetic progression. The answer is 180. Here's the breakdown:
Understanding Arithmetic Progressions and Divisibility by 5
An arithmetic progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. A number is divisible by 5 if its last digit is either 0 or 5.
Deriving the Solution
Let the three-digit numbers in arithmetic progression be a - d, a, and a + d, where a is the middle term and d is the common difference. All three numbers must be divisible by 5 and fall within the range of 100 to 999.
Since a - d, a, and a + d are all divisible by 5, their common difference, d, must also be divisible by 5. Furthermore, since a is a three-digit number divisible by 5, the smallest possible value for a is 100, and the largest possible value is 995.
The general form of such a sequence is a - d, a, a + d. Since all terms must be three-digit numbers:
- 100 ≤ a - d
- 100 ≤ a ≤ 999
- a + d ≤ 999
We know a must be divisible by 5.
Let's find the possible values of d. Since the smallest three-digit number divisible by 5 is 100, and the largest is 995, consider the sequence where a = 100. The common difference d can vary. If d is large, a - d could become a two-digit number and a + d could become a four-digit number. The same logic applies when a = 995.
Here's a simpler way to approach the problem: Since a - d, a, and a + d are all divisible by 5, we can express them as 5x, 5y, and 5z for some integers x, y, and z. The condition for an arithmetic progression is:
5y - 5x = 5z - 5y which simplifies to y - x = z - y or 2y = x + z.
Therefore, all three terms of the sequence must be multiples of 5. The number of three digit numbers divisible by 5 is calculated as follows:
- The smallest three-digit number divisible by 5 is 100.
- The largest three-digit number divisible by 5 is 995.
We can express the sequence of three-digit numbers divisible by 5 as: 100, 105, 110, ..., 995. The nth term of this AP is given by:
an = a1 + (n-1)d, where a1 = 100, d = 5, and an = 995
995 = 100 + (n-1)5
895 = (n-1)5
179 = n - 1
n = 180
Therefore, there are 180 three-digit numbers that are divisible by 5. Since each of these numbers can be the central element ('a') of an arithmetic progression where all three elements are three-digit and divisible by 5, there are 180 possible arithmetic progressions.
Conclusion
Therefore, there are 180 possible arithmetic progressions of three-digit numbers divisible by 5.
