how do you solve this algebra problem? arithmetic series?

Question

if the first term is -4 and the common difference is 4, the sum is 80. find the number of terms in this finite arithmetic series.
how do u do this and whats the answer?

Answer 1

There is a formula for S(n), the sum of the first n terms of an arithmetic sequence:

S(n) = .5n (2a1 + d(n-1))

You have S(n), a1 and d, but you do not have n. But we can solve for n as follows:

80 = .5n(2*-4 + 4(n-1))
160 = n(-8 + 4n-4)
160 = n(4n-12)
40 = n(n-3)
n² - 3n - 40 = 0
(n-8)(n+5) = 0
n=8 or -5 but n = -5 is not valid for the problem since n>0.

Answer 2

a[0] = -4
a[n] - a[n-1] = 4 ; n>=1
sum a[n] = 80.

how i would do this calculate the sum entil the sum is 80.

a1=-4 sum=-4
a2=0 sum=-4
a3=4 sum=0
a4=8 sum = 8
a5=12 sum = 20
a6=16 sum = 36
a7=20 sum = 56
a8=24 sum = 80

Answer 3

I would say that it's probably about 30. It may be 29, I'm not sure though.


Truthfully?

An = a1 + (n-1)d

d= common difference
a1 = first term

So, we do:

-4, 0, 4, 8, etc.

80 = -4 + (n-1)(4)
80 = -4 + 4n - 4
80 = 4n - 8
88 = 4n
n = 22

There are 22 numbers, k babe?

Answer 4

if the first 8 terms are

-4,0,4,8,12,16,20,24 then
Their sum is 80