How many terms of the series 25+19+13 +… are required to make the sum −20?

Question

How many terms of the series 25+19+13 +… are required to make the sum −20?

Answer 1

sum of arithmetic series = n(2a(1)+ (n-1)d)/2

a(1) = 25

d = -6

sum = -20

-20 = n(50 + (n-1)-6)/2

-40 = n(50 - 6n + 6)

-40 = 50n - 6n^2 + 6n

-40 = -6n^2 + 56n

6n^2 -56n - 40 = 0

3n^2 -28n - 20 = 0

3n^2 - 30n + 2n - 20 = 0

3n(n - 10) + 2(n-10) = 0

(3n + 2)(n-10) = 0

n = 10

Answer 2

Use the formula for sum of an arithmetic series:

S_n = (n/2)(t₁ + t_n)

(n/2) (25 + (25 - 6(n - 1))) = - 20

n(56 - 6n) = - 40

6n² - 56n - 40 = 0

2(3n + 2)(n - 10) = 0

n = 10