S_n = \fracn2 \left(2a + (n - 1)d\right) = \fracn2 \left(2(7) + (n - 1)(4)\right) = \fracn2 (14 + 4n - 4) = \fracn2 (4n + 10) - Nelissen Grade advocaten
S_n Formula Explained: Mastering the $n$-th Term of an Arithmetic Sequence
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:
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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
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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:
