91Ó°ÊÓ

In an arithmetic sequence, finding the sum of its terms is a common task. When dealing with the sum of multiple terms, particularly when they form a sequence like this one, it's easier to use the arithmetic series formula. This formula is designed to help us quickly find the total of a specific number of terms in such a sequence. For example, if you need the sum of the first 50 terms, accumulating each term manually would be cumbersome. Instead, you can use the series formula which is:

• \( S_n = \frac{n}{2} (a_1 + a_n) \)

Here, \( S_n \) represents the sum of the first \( n \) terms where \( n \) is the total number of terms you need to sum up. \( a_1 \) is the first term in the sequence, and \( a_n \) is the \( n \)-th term of the sequence. By utilising this method, calculating the sum becomes much simpler and quicker.

Understanding the common difference is crucial when dealing with arithmetic sequences. The common difference is the fixed amount by which each term in the sequence increases from the previous one. Identifying this consistent difference helps in determining any term of the sequence without having to enumerate all preceding terms.
In our example, the sequence starts at \(-10\) and proceeds to \(-6\), then \(-2\), and so on. Here, the common difference \( d \) is 4, because each term increases by 4 from its predecessor. Knowing this value, you can accurately find any specific term in the sequence using the formula for the nth term:

• \( a_n = a_1 + (n-1) \times d \)

Again, once you know \( a_1 \) and \( d \), you can quickly compute terms or employ these in formulas for solving for terms or sums.

The arithmetic series formula is a powerful tool for finding the sum of sequences composed of terms regularly spaced by a common difference. It saves time and effort by providing a direct way to compute this sum without manual addition of each term. The formula used is:

• \( S_n = \frac{n}{2} (a_1 + a_n) \)

This formula can be derived from the known properties of arithmetic sequences. It accounts for every member from the first term to the nth term by calculating an average term and multiplying it by their count, \( n/2 \).
For instance, instead of manually adding every term from \(-10\) to 186 in our problem, the formula takes this workload down to strategized arithmetic:- Add the first and the nth terms, \(-10 + 186\)- Multiply by half the number of terms (50/2 = 25)- Yielding a swift calculation for the total sum of 4400, showcasing its efficiency and reliability.