✎ Answer
12 terms must be in the sequence to sum up to 120.
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✎ Solution:
We are given the following values:
- common difference d : 2
- first term a₁: -1
- sum of the terms Sₙ: 120
We are required to find the number of terms n. So we will use the formula for the arithmetic series.
Sₙ = n/2 [2a₁ + (n - 1) d]
We plug in the values to solve for n.
- 120 = n/2 [2(-1) + (n - 1)(2)]
- 120 = n/2 [-2 + (2n - 2)]
- 120 = n/2 (2n - 4)
- (2)(120) = [n/2 (2n - 4)] 2
- 240 = n (2n - 4)
- 240 = 2n² - 4n
- 2n² - 4n - 240 = 0
- 2 (n² - 2n - 120) = 0
- 2 [n² + (-12n + 10n) - 120] = 0
- 2 [(n² - 12n) + (10n - 120)] = 0
- 2 [n (n - 12) + 10 (n - 12)] = 0
- 2 [(n + 10)(n - 12)] = 0
We have to equate each of the factors to zero to find the value of n:
- n + 10 = 0; n = -10
- n - 12 = 0; n = 12
It is apparent that n must be a positive integer, thus n = 12 is the answer.
✎ Check:
Let's check if the number of terms of the given sequence sum up to 120, using the formula for the arithmetic series:
- Sₙ = n/2 [2a₁ + (n - 1) d]
- 120 = (12/2)[2(-1) + (12 - 1)(2)]
- 120 = (6)[(-2) + (22)]
- 120 = (6)[(20)]
- 120 = 120
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