How many terms are added in the arithmetic sequence –1, 1, 3, 5, ... if its partial sum is 80?

The arithmetic sequence starts with -1 with a common difference of 2.

The formula for the partial sums of an arithmetic sequence is:

Sn = n/2 * (2a + d(n-1))

Sn : sum of first n terms (80)

a : first term (-1)

d : common difference (2)

n : number of terms (unknown)

Plug everything in and solve for n:

80 = n/2 * (2(-1) + 2(n - 1))

80 = n/2 * (-2 + 2n - 2)

80 = n/2 * (2n - 4)

80 = n(n - 2)

80 = n² - 2n

n² - 2n - 80 = 0

Factor that:

(n - 10)(n + 8) = 0

Using the zero product rule:

n - 10 = 0

n = 10

or

n + 8 = 0

n = -8

We know we need a positive number of terms so we can eliminate -8 as an answer.

Let's double-check:

-1 + 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 80

Answer:

10 terms

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