Arithmetic Series: Practice

A sequence is a list of terms, while a series is what you get by adding those terms together.

$2 + 5 + 8 + 11 = 26$

The plus signs make it a sum, so it is a series; the others are lists.

A sequence is the ordered list; a series adds the entries of that list.

$S_n = a_1 + a_2 + \cdots + a_n$

So a series is the sum of the terms of a sequence.

Substitute $k = 1, 2, 3, 4$ into the summand $2k$ and add.

$2 + 4 + 6 + 8 = 20$

The terms run from position $1$ to position $12$, so count $12 - 1 + 1$.

$12 - 1 + 1 = 12$

There are $12$ terms.

Each term is $3$ times its position, so the summand is $3k$, and there are $5$ terms.

$\sum_{k=1}^{5} 3k = 3 + 6 + 9 + 12 + 15$

Use $S_n = \frac{n(a_1 + a_n)}{2}$ with the first term, last term, and count given.

$S_{10} = \frac{10(2 + 20)}{2} = \frac{220}{2} = 110$

Adding the terms of a sequence gives a series; the sum of the first $n$ of them is a partial sum.

$S_n = a_1 + a_2 + \cdots + a_n$

That total $S_n$ is the (partial) sum, a series.

Use $S_n = \frac{n(a_1 + a_n)}{2}$ with $n = 5$, $a_1 = 4$, and $a_5 = 16$.

$S_5 = \frac{5(4 + 16)}{2} = \frac{5 \times 20}{2} = 50$

Substitute $k = 1, 2, 3, 4$ into the summand $k + 2$.

$(1+2) + (2+2) + (3+2) + (4+2) = 3 + 4 + 5 + 6$

The sum is the count times the average of the first and last term.

$S_5 = 5 \times \frac{2 + 10}{2} = 5 \times 6 = 30$

Adding directly, $2 + 4 + 6 + 8 + 10 = 30$, agrees.

Use $S_n = \frac{n(a_1 + a_n)}{2}$ with $n = 20$, $a_1 = 1$, and $a_{20} = 20$.

$S_{20} = \frac{20(1 + 20)}{2} = \frac{20 \times 21}{2} = 210$

Use $S_n = \frac{n(a_1 + a_n)}{2}$ with $n = 4$, $a_1 = 10$, and $a_4 = 40$.

$S_4 = \frac{4(10 + 40)}{2} = \frac{4 \times 50}{2} = 100$