Precal Help?

Question

In a certain arithmetic sequence, a = −5 and d = 6. If Sn = 476, find n.

n =

Answer 1

In arithmetic sequence, the sum of n terms : S_n = (n/2)[2a + (n - 1)d].

Where, a is the first term of the arithmetic sequance,

n is the number of terms,

d is the common difference.

Given : a = - 5, d = 6, and S_n = 476.

Substitute the corresponding values in sum formula : S_n = (n/2)[2a + (n - 1)d].

476 = (n/2)[2(- 5) + (n - 1)6]

476 * 2 = n[- 10 + 6n - 6]

476 * 2 = n[6n - 16]

476 * 2 = 2[3n² - 8n]

476 = 3n² - 8n

3n² - 8n - 476 = 0

By factor by grouping.

3n² - 42n + 34n - 476 = 0

3n(n - 14) + 34(n - 14) = 0

(n - 14)(3n + 34) = 0

Apply zero product property.

n - 14 = 0  and  3n + 34 = 0

n = 14  and  n = - 34/3

n = - 34/3 is not possible, n must be a positive integer.

So, the number of terms in the arithmetic sequence(n) is 14.