Question

What is the sum of the multiples of 4 between 13 and 125 inclusive? a. 1,890 b. 1,960 c. 2,200 d. 3,780 e. 4,400

Short answer

The sum of the multiples of 4 between 13 and 125 is 1,960.

Step-by-step solution

- Identify the multiples of 4 in the given range

The multiples of 4 within the range of 13 to 125 are: 16, 20, 24, ..., 124. These can be identified by finding the smallest multiple of 4 greater than or equal to 13, and the largest multiple of 4 less than or equal to 125.

- Determine the sequence

The sequence of multiples of 4 between 13 and 125 forms an arithmetic sequence. The first term (a) is 16, and the common difference (d) is 4.

- Find the last term

The largest multiple of 4 within the range is 124.

- Calculate the number of terms (n) in the sequence

Use the formula for the nth term of an arithmetic sequence: \[ a_n = a + (n-1)d \]. Setting \( a_n \) to 124, we solve for n:\[ 124 = 16 + (n-1) \times 4 \]\[ 124 = 16 + 4n - 4 \]\[ 124 = 12 + 4n \]\[ 112 = 4n \]\[ n = 28 \]. Thus, there are 28 terms in the sequence.

- Sum the arithmetic sequence

The sum \( S_n \) of an arithmetic sequence can be calculated using the formula: \[ S_n = \frac{n}{2} (a + l) \] where \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term. Substituting the known values: \[ S_{28} = \frac{28}{2} (16 + 124) \]\[ S_{28} = 14 \times 140 \]\[ S_{28} = 1960 \].

- Verify the answer

The sum is 1960, so the correct choice is b. 1,960.

Key concepts

Understanding Multiples

Identifying multiples is the first essential step in solving problems about arithmetic sequences of multiples. Multiples of a number are the results you get when you multiply that number by integers. For example, the multiples of 4 are obtained by multiplying 4 by 1, 2, 3, and so on.
In this problem, we identified that the multiples of 4 between 13 and 125 are: 16, 20, 24, ..., 124.
Finding these multiples involves checking for the smallest multiple of 4 that is greater than or equal to 13, and the largest multiple of 4 that is less than or equal to 125.
Thus, you can see that the identified multiples create a specific sequence we need to analyze.

Sum of Sequence

The sum of a sequence, especially an arithmetic sequence (where there is a consistent difference between terms), can be calculated efficiently using a particular formula: \[ S_n = \frac{n}{2} (a + l) \], where \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term.
In this exercise, once we established that we have 28 terms, starting at 16 and ending at 124, we can easily use this sum formula.
Substituting our values: \ S_{28} = \frac{28}{2} (16 + 124) \
This arithmetic progression allows us to find: \ S_{28} = 14 \times 140 = 1960 \
which, as verified, results in the correct answer.

Calculating Number of Terms

To determine the number of terms in an arithmetic sequence, we use the formula for the nth term, which is: \ a_n = a + (n-1) \times d \.
Here, \( a \) is the first term (16 in our sequence), \( d \) is the common difference (4), and \( a_n \) is the last term (124).
We set up the equation: \ 124 = 16 + (n-1) \times 4 \
Solving this step-by-step: \ 124 = 16 + 4n - 4 \
\ 124 = 12 + 4n \
\ 112 = 4n \
\ n = 28 \
This tells us that there are 28 terms in this sequence.
It's always crucial to check your solution twice to avoid any mistakes in your calculations.

Common Difference

The common difference in an arithmetic sequence is the difference between consecutive terms. It is a key element that defines the sequence.
For the set of multiples of 4, our common difference is 4 because each term increases by 4 from the previous term.
Recognizing the common difference helps justify the sequence formation and aids in the application of arithmetic formulas.
When solving for the sum or identifying the number of terms in a sequence, knowing the common difference is essential.
Always verify that the difference remains consistent throughout the sequence to confirm it is indeed an arithmetic sequence.
In summary, the common difference simplifies calculations, ensuring the correct progression of terms.