S_n Formula Explained: Mastering the $n$-th Term of an Arithmetic Sequence

When studying mathematics, especially algebra and sequences, one formula emerges as essential for finding the $n$-th term of an arithmetic sequence:

$$
S_n = \frac{n}{2} \left(2a + (n - 1)d\right)
$$

Understanding the Context

This elegant expression allows you to compute the sum of the first $n$ terms of any arithmetic sequence quickly — without having to add every term individually.

Understanding the Formula

The formula
$$
S_n = \frac{n}{2} \left(2a + (n - 1)d\right)
$$
is the standard formula for the sum of the first $n$ terms ($S_n$) of an arithmetic sequence, where:

Solved f\left(x\right)=\frac{1}{x},\:1:1:\left(1,1:3\right] | Chegg.com

Image Gallery

Solved f\left(x\right)=\frac{1}{x},\:1:1:\left(1,1:3\right] | Chegg.com
prove by induction that for all n 1 47 3n 2 2 n3n 1 iii 246zn nn 1 6 12 ...
Solved Use mathematical induction to prove that 2 - (2)(7) + | Chegg.com
sequences and series - $\sum _{k=1}^n\:\left(\cos\left(\frac{2\cdot k ...
sequences and series - Evaluating $\sum_{n=0}^{\infty} \left(\frac{1}{2 ...

Key Insights

  • $S_n$ = sum of the first $n$ terms
    - $a$ = the first term of the sequence
    - $d$ = common difference between consecutive terms
    - $n$ = number of terms to sum

It is derived from pairing terms in reverse order:
$ a + (a + d) + (a + 2d) + \cdots + [a + (n - 1)d) $

Pairing the first and last terms gives $a + [a + (n - 1)d] = 2a + (n - 1)d$, and with $n$ such pairs multiplied by $\frac{n}{2}$, we get the formula above.

Plugging in Sample Values

Continue Reading

Question:** A nanotechnologist is analyzing the behavior of a nanoparticle where the resistance \( R \) is modeled by \( R = \frac{4}{\sqrt{x} + 2} \). Rationalize the denominator of the expression for \( R \).
To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator:
R = \frac{4}{\sqrt{x} + 2} \times \frac{\sqrt{x} - 2}{\sqrt{x} - 2}

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Final Thoughts

Let’s analyze the specific case given in the formula:

$$
S_n = \frac{n}{2} \left(2(7) + (n - 1)(4)\right) = \frac{n}{2} (14 + 4n - 4) = \frac{n}{2} (4n + 10)
$$

Here:
- $a = 7$
- $d = 4$

So the sequence begins:
$7, 11, 15, 19, \ldots$
Each term increases by $4$. Using the sum formula gives a fast way to compute cumulative sums.

For example, find $S_5$:

$$
S_5 = \frac{5}{2} (4 \cdot 5 + 10) = \frac{5}{2} (20 + 10) = \frac{5}{2} \ imes 30 = 75
$$

Indeed, $7 + 11 + 15 + 19 + 23 = 75$, confirming the formula’s accuracy.

Why This Formula Matters

The $S_n = \frac{n}{2}(2a + (n - 1)d)$ formula is indispensable in: