Find the 7th partial sum of the sequence {a_n} = -9 + 3(n – 1).

We're gonna need the first term and common difference. Where you at, fellas?

a₁ = -9

d = 3

Now find the 7th term of the sequence using the explicit rule provided.

{a_n} = -9 + 3(n – 1)
a₇ = -9 + 3(7 – 1)
a₇ = -9 + 3(6)
a₇ = -9 + 18
a₇ = 9

Plug the first term, 7th term, and n into the partial sum formula.

Simplify that bad boy.

Whoa, our sum is just 0. All the negative terms and positive terms canceled each other out perfectly.

Find the 14th partial sum of  the sequence {a_n} = -4n + 20.

All we need to do to evaluate this partial sum is to find the number of terms as well as the first and last terms. Let's start by listing the first few terms to find the first term and common difference, d.

a₁ = -4(1) + 20 = 16
a₂ = -4(2) + 20 = 12
a₃ = -4(3) + 20 = 8

Rock on. Our first term is a₁ = 16, and since we're subtracting 4 from each term, the common difference is d = -4. 

Now we use the given rule to find the 14th term by plugging 14 in for n.

a₁₄ = -4(14) + 20
a₁₄ = -36

Sum time. Use the formula for the partial sum of an arithmetic series.

Now plug everything in and simplify to find your final solution.

Find the sum of -6, 0, 6,…, 66, 72.

As usual, we'll need the first term, last term, and common difference.

a₁ = -6

a_n = 72

d = 6

Plug these dudes into the explicit rule for the sequence and solve for n, the number of terms in the sequence.

72 = -6 + 6(n – 1)
78 = 6n – 6
84 = 6n
14 = n

Cool, so we've got 14 terms in our sequence. Now bust out the partial sum formula.

Plug in a₁ = -6, a_n = 72, and n = 14.