# Arithmetic Series

• Nov 24th 2009, 08:51 PM
Arithmetic Series
I'm trying to find the number of terms in the arithmetic series $-4, 2, 8, 14, ....$ which makes the sum strictly greater than $18,000$.

I know that $S_{n} = \frac{1}{2}\left[{2a+(n-1)d}\right]$, but I somehow can't apply it to this problem.
• Nov 24th 2009, 08:53 PM
Jameson
Quote:

I'm trying to find the number of terms in the arithmetic series $-4, 2, 8, 14, ....$ which makes the sum strictly greater than $18,000$.

I know that $S_{n} = \frac{1}{2}\left[{2a+(n-1)d}\right]$, but I somehow can't apply it to this problem.

You know a_1 and d. Set S_n=18,000 and solve for n.
• Nov 24th 2009, 09:16 PM
Quote:

Originally Posted by Jameson
You know a_1 and d. Set S_n=18,000 and solve for n.

$a = -4$

$d = 6$

$\frac{1}{2}\left[2(-4)+(n-1)6\right] = 18000$

$\frac{1}{2}\left[-8+6n-6\right] = 18000$

$\frac{1}{2}\left[-8+6n-6\right] = 18000$

$\frac{1}{2}\left[6n-14\right] = 18000
$

3 $n-7 = 18,000$

$3n = 18,000 -7$

$3n = 17993$

$n = \frac{17993}{3}$

$n = 5997.666667$

$5998$ terms?
• Nov 24th 2009, 09:46 PM
Jameson
Almost. You subtracted 7 but should have added.
• Nov 24th 2009, 09:50 PM
Quote:

Originally Posted by Jameson
Almost. You subtracted 7 but should have added.

Oh, silly mistake.

$3n = 18,000 +7$

$3n = 18007$

$n = \frac{18007}{3}$

$n = 6002.333333$

$6003$ terms then?
• Nov 24th 2009, 09:56 PM
Jameson
Seems to make sense. (Yes)
• Nov 24th 2009, 10:00 PM