Understanding the Sum of the First 10 Terms Using the Arithmetic Series Formula

Calculating the sum of a sequence is a fundamental concept in mathematics, especially when working with arithmetic series. One interesting example involves computing the sum of the first 10 terms of a specific series using a well-known formula — and it aligns perfectly with \( \frac{10}{2} \ imes (5 + 32) = 5 \ imes 37 = 185 \). In this article, we’ll explore how this formula works, why it’s effective, and how you can apply it to solve similar problems efficiently.

Understanding the Context

What Is an Arithmetic Series?

An arithmetic series is the sum of the terms of an arithmetic sequence — a sequence in which the difference between any two consecutive terms is constant. This constant difference is called the common difference.

For example, consider the sequence:
\( a, a + d, a + 2d, a + 3d, \dots \)
where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the number of terms.

The sum of the first \( n \) terms of such a series is given by the formula:

The sum of first 10 terms of the series \( \left(x+\frac{1}{x}\right ...

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The sum of first 10 terms of the series \( \left(x+\frac{1}{x}\right ...
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Key Insights

\[
S_n = \frac{n}{2} \ imes (2a + (n - 1)d)
\]

Alternatively, it can also be written using the average of the first and last term:

\[
S_n = \frac{n}{2} \ imes (a + l)
\]

where \( l \) is the last term, and \( l = a + (n - 1)d \).

Continue Reading

Let \(h(t) = \frac{1 + 2t}{t^2}\), \(t \in (0, \frac{1}{2}]\)
As \(t \to 0^+\), \(h(t) \to \infty\), and \(h'(t) = \frac{2t^2 - (1 + 2t)(2t)}{t^4} = \frac{2t - 2(1 + 2t)}{t^3} = \frac{2t - 2 - 4t}{t^3} = \frac{-2t - 2}{t^3} < 0\), so decreasing.
Maximum value is approached as \(t \to 0^+\), i.e., as \(x \to 0^+\) or \(\frac{\pi}{2}^-\), and \(

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Final Thoughts

Applying the Formula to the Given Example

In the expression:
\[
\frac{10}{2} \ imes (5 + 32) = 5 \ imes 37 = 185
\]
we recognize this as a concise application of the arithmetic series sum formula.

Let’s match the terms to our general strategy:

  • \( n = 10 \) — we want the sum of the first 10 terms
    - First term, \( a = 5 \)
    - Last term, \( l = 32 \) (which is \( 5 + (10 - 1) \ imes d = 5 + 9 \ imes d \). Since \( d = 3 \), \( 5 + 27 = 32 \))

Now compute:
\[
S_{10} = \frac{10}{2} \ imes (5 + 32) = 5 \ imes 37 = 185
\]

Why This Formula Works

The formula leverages symmetry in the arithmetic series: pairing the first and last terms, the second and second-to-last, and so on until the middle. Each pair averages to the overall average of the sequence, \( \frac{a + l}{2} \), and there are \( \frac{n}{2} \) such pairs when \( n \) is even (or \( \frac{n - 1}{2} \) pairs plus the middle term when \( n \) is odd — but not needed here).

Thus,

\[
S_n = \frac{n}{2} \ imes (a + l)
\]