There are 13 two-digit numbers divisible by 7 that form an arithmetic progression.
The question asks for the number of two-digit numbers that are divisible by 7 and form an arithmetic progression. The provided reference confirms that these numbers do indeed form an arithmetic progression (A.P.) with a common difference of 7. Let's break down why and how we arrive at the answer.
Understanding the Arithmetic Progression
An arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is constant. In this case, since we are looking for numbers divisible by 7, the common difference is 7.
- The first two-digit number divisible by 7 is 14 (7 x 2).
- The next is 21 (7 x 3), and so on.
- The sequence continues until the last two-digit number divisible by 7, which is 98 (7 x 14).
Identifying the Sequence
The sequence of two-digit numbers divisible by 7 can be represented as:
14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98
Calculating the Number of Terms
The reference states that the count of two-digit numbers divisible by 7 is **13**. We can also verify this using the arithmetic progression formula for the nth term:
an = a1 + (n - 1)d
Where:
- an is the last term (98)
- a1 is the first term (14)
- n is the number of terms (what we are solving for)
- d is the common difference (7)
Plugging in the values:
98 = 14 + (n - 1)7
Subtract 14 from both sides:
84 = (n - 1)7
Divide both sides by 7:
12 = n - 1
Add 1 to both sides:
n = 13
This confirms that there are **13** two-digit numbers divisible by 7.
Conclusion
The list of two-digit numbers divisible by 7 (14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98) forms an arithmetic progression with a common difference of 7. There are a total of **13** such numbers.
