To find the sum of all the numbers between 100 and 200 which are divisible by 7, we can use the formula for the sum of an arithmetic series. An arithmetic series is a sequence of numbers in which each term after the first is obtained by adding a constant difference to the preceding term. Since the difference between numbers divisible by 7 is 7 (the common difference 'd'), this is an arithmetic series.
Here's how we solve this problem step-by-step:
1. Determine the first number greater than 100 that is divisible by 7.
To find this number, we can take the remainder when 100 is divided by 7 (100 modulo 7), and subtract that from 7 to find the amount we need to add to 100 to get to the next multiple of 7:
\[ 100 \mod 7 = 2 \]
So, we need to add \( 7 - 2 = 5 \) to 100 to get the first number greater than 100 divisible by 7:
\[ 100 + 5 = 105 \]
2. Determine the last number less than 200 that is divisible by 7.
To find this number, we simply take 200 and subtract the remainder when 200 is divided by 7 (200 modulo 7):
\[ 200 \mod 7 = 4 \]
So, we subtract 4 from 200 to get the last number less than 200 divisible by 7:
\[ 200 - 4 = 196 \]
3. Now we need to find the sum of this series. But first, we need to determine how many terms are there in this series. We can use the general formula to find the nth term of an arithmetic series:
\[ a_n = a_1 + (n - 1)d \]
Where:
\[ a_n \] is the last term (which we found to be 196),
\[ a_1 \] is the first term (which we found to be 105), and
\[ d \] is the common difference (which is 7 in this case).
Rearranging the formula to solve for n, we get:
\[ n = \frac{(a_n - a_1)}{d} + 1 \]
Substituting the known values, we have:
\[ n = \frac{(196 - 105)}{7} + 1 \]
\[ n = \frac{91}{7} + 1 \]
\[ n = 13 + 1 \]
\[ n = 14 \]
So, there are 14 terms in the series.
4. Finally, we apply the sum formula for an arithmetic series which is given by:
\[ S = \frac{n}{2} \times (a_1 + a_n) \]
Where:
\[ S \] is the sum of the series,
\[ n \] is the number of terms (14, as we found above),
\[ a_1 \] is the first term (105), and
\[ a_n \] is the last term (196).
Now we substitute these values into the formula:
\[ S = \frac{14}{2} \times (105 + 196) \]
\[ S = 7 \times 301 \]
\[ S = 2107 \]
Therefore, the sum of all the numbers between 100 and 200 that are divisible by 7 is 2107.
