Sum of Even Multiples of 3 Between 100 and 250: A Comprehensive Guide
This guide provides a detailed step-by-step approach to finding the sum of even multiples of 3 between 100 and 250. The process involves identifying the range of even multiples, determining the smallest and largest even multiples, listing the sequence, and finally, calculating the sum of this arithmetic sequence.
Identifying the Range of Even Multiples of 3
Even multiples of 3 are multiples of 6, as 6 is the product of 2 and 3. The multiples of 3 follow the form 3n, where n is an integer. To find the even multiples, we use the form 6n.
Step 1: Finding the Smallest Even Multiple of 3 Greater Than or Equal to 100
We use the inequality:
n ≥ 100/6 ≈ 16.67
The smallest integer n that satisfies this condition is 17. Therefore, the smallest even multiple of 3 is:
6 × 17 102
Step 2: Finding the Largest Even Multiple of 3 Less Than or Equal to 250
Similarly, we solve the inequality:
n ≤ 250/6 ≈ 41.67
The largest integer n that satisfies this condition is 41. Therefore, the largest even multiple of 3 is:
6 × 41 246
Listing the Even Multiples of 3 from 102 to 246
The even multiples of 3 from 102 to 246 form an arithmetic sequence where:
The first term a 102 The last term l 246 The common difference d 6To find the number of terms in this sequence, we use the formula for the nth term of an arithmetic sequence:
l a (n-1)d
Rearranging gives:
n (l - a)/d 1 (246 - 102)/6 1 144/6 1 24 1 25
Thus, there are 25 even multiples of 3 between 102 and 246.
Calculating the Sum of the Arithmetic Sequence
The sum of the first n terms of an arithmetic sequence can be calculated using the formula:
Sn (n/2)(a l)
Substituting the values we found:
S25 (25/2)(102 246) (25/2) × 348 25 × 174 4350
Therefore, the sum of the even multiples of 3 between 100 and 250 is 4350.
For a verification, the sequence is:
102, 108, 114, ..., 246
This is an arithmetic progression (AP) with a common difference of 6. The nth term of an AP is given by:
an a (n-1)d
For the last term 246:
246 102 (n-1) × 6
Solving for n:
(n-1) × 6 246 - 102 144
n-1 144/6 24
n 25
The sum of the first n terms of an AP is given by:
Sn n/2 × (first term last term)
Thus, S25 25/2 × (102 246) 25/2 × 348 25 × 174 4350
Conclusion
By following these steps, we can systematically find the sum of the even multiples of 3 between 100 and 250. The sum is 4350, as demonstrated through both direct arithmetic and the properties of an arithmetic sequence.