Arithmetic series question?

Question

Here is an arithmetic series:
4 + 13 + 22 + 31 + ...
If the sum of this arithmetic series is to be greater than 1000, how many terms are needed to achieve this?

Show me how you work this out.

Answer 1

Best Answer - Chosen by Asker

t(n) = 9n - 5

s(n) = sum t(n)

= sum(9n) - 5 sum(1)

= 9 sum(n) - 5 sum(1)

= 9n(n+1)/2 - 5n

= (9/2) n² - (1/2) n

> 1000

quadratic equation

(9/2)n² - (1/2)n - 1000 = 0

disc : (1/2)² + 4(9/2)(1000) = 18000.25

sqrt(disc) = 134.16501

n = ((1/2) +- 134.16501...) / 9

= 14.962779 (n>0)

so we need 15 terms

Asker's Rating:

Asker's Comment:

thanks for clearing that up

Answer 2

I forgot the formula....but u mite still be able to do it
take 1000 and add 4 to it because thats the # u started with
then divide 1004 by 9 = 111.555.....6
then round it so you get 112 and hopefully thats how many terms r needed

Answer 3

Umm well the way I learned it was:

4n + 13n + 22n + 31n = 1000

70n = 1000
---- ------- (divide each side by 70)
70 70
_________
n ~
~ 15 terms

so about 15 terms