There are 30 integers between 10 and 100 that are divisible by 3.
The integers divisible by 3 between 10 and 100 form an arithmetic progression (AP). This means they follow a sequence where the difference between consecutive terms is constant.
Here's a breakdown:
- First term (a): The first integer greater than 10 that is divisible by 3 is 12.
- Common difference (d): The difference between consecutive multiples of 3 is 3.
- Last term: The last integer less than 100 that is divisible by 3 is 99.
Let's represent the number of integers in this sequence as 'n'. We can find 'n' using the formula for the nth term of an AP:
Last Term = a + (n - 1) * d
Plugging in the values we have:
99 = 12 + (n - 1) * 3
Solving for n:
99 - 12 = (n - 1) 3
87 = (n - 1) 3
87 / 3 = n - 1
29 = n - 1
n = 30
Therefore, there are 30 integers between 10 and 100 that are divisible by 3. The reference states, "These integers form an AP where first term = a = 12, common difference = d = 3, last term = 99. Let n be the number of these integers. = n = 30. Hence there are 30 numbers divisible by 3 from 10 and 100.09-Mar-2020".
