Finding the Sum of Numbers Divisible by 12 Between 100 and 200

The problem asks us to find the sum of all whole numbers that are greater than 100 and less than 200, and are also perfectly divisible by 12.

The numbers divisible by 12 form an arithmetic progression. We need to identify the first term, the last term, and the common difference of this progression that falls within the specified range (100 to 200).

Step-by-Step Solution to Find the Sum

Let's break down the process:

Step 1: Identify the First Number Divisible by 12 Between 100 and 200.

We need to find the smallest multiple of 12 that is greater than 100. Let's divide 100 by 12:

\( \frac{100}{12} \approx 8.33 \)

This means 100 is between the 8th and 9th multiple of 12. The multiples of 12 are 12, 24, 36, ..., 96 (\(12 \times 8\)), 108 (\(12 \times 9\)), and so on.

The smallest multiple of 12 greater than 100 is \(12 \times 9 = 108\). So, the first term of our arithmetic progression is 108.

Step 2: Identify the Last Number Divisible by 12 Between 100 and 200.

We need to find the largest multiple of 12 that is less than 200. Let's divide 200 by 12:

\( \frac{200}{12} \approx 16.67 \)

This means 200 is between the 16th and 17th multiple of 12. The multiples of 12 continue past 108: 120, 132, ..., \(12 \times 16 = 192\), \(12 \times 17 = 204\), etc.

The largest multiple of 12 less than 200 is \(12 \times 16 = 192\). So, the last term of our arithmetic progression is 192.

Step 3: Determine the Arithmetic Progression.

The numbers between 100 and 200 that are divisible by 12 form an arithmetic progression starting at 108, ending at 192, and with a common difference of 12. The sequence is: 108, 120, 132, 144, 156, 168, 180, 192.

Here, we have:

• First term (\(a\)) = 108
• Common difference (\(d\)) = 12
• Last term (\(l\)) = 192

Step 4: Find the Number of Terms in the Progression.

We can find the number of terms (\(n\)) using the formula for the \(n\)-th term of an arithmetic progression: \(l = a + (n-1)d\).

Substitute the values:

\( 192 = 108 + (n-1) \times 12 \)

Subtract 108 from both sides:

\( 192 - 108 = (n-1) \times 12 \)

\( 84 = (n-1) \times 12 \)

Divide both sides by 12:

\( \frac{84}{12} = n-1 \)

\( 7 = n-1 \)

Add 1 to both sides:

\( n = 7 + 1 \)

\( n = 8 \)

There are 8 numbers between 100 and 200 that are divisible by 12.

Step 5: Calculate the Sum of the Arithmetic Progression.

We can use the formula for the sum of an arithmetic progression when the first and last terms are known: \( S_n = \frac{n}{2}(a + l) \).

Substitute the values:

\( S_8 = \frac{8}{2}(108 + 192) \)

\( S_8 = 4(300) \)

\( S_8 = 1200 \)

The sum of all numbers between 100 and 200 that are divisible by 12 is 1200.

Summary of the Calculation

Therefore, the sum of the numbers between 100 and 200 which are divisible by 12 is 1200.

Revision Table: Key Concepts for Divisibility and Arithmetic Series

Additional Information on Finding Sums of Divisible Numbers

Problems asking for the sum of numbers divisible by a specific value within a range are common applications of arithmetic progressions. The key steps are always:

• Identify the first term in the range that satisfies the divisibility condition.
• Identify the last term in the range that satisfies the divisibility condition.
• Confirm the common difference is the divisor itself.
• Calculate the number of terms in the sequence.
• Use the sum formula for an arithmetic progression.

Remember that "between 100 and 200" usually means strictly greater than 100 and strictly less than 200. If the question included "from 100 to 200", we would check if 100 or 200 themselves were divisible by 12 and include them if they were.