Arithmetic Series

**Captcha** · November 24th 2009, 07:51 PM

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.

**Jameson** · November 24th 2009, 07:53 PM

Originally Posted by Captcha

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.

**Captcha** · November 24th 2009, 08:16 PM

Originally Posted by Jameson

You know a_1 and d. Set S_n=18,000 and solve for n.

$a = -4$

$d = 6$

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

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

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

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

3 $n-7 = 18,000$

$3n = 18,000 -7$

$3n = 17993$

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

$n = 5997.666667$

$5998$ terms?