# Find the least value of n in an arithmetic sequence

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• Oct 18th 2009, 09:35 AM
Viral
Ok, I've used that formula but don't know how to solve for n.
• Oct 18th 2009, 09:42 AM
Quote:

Originally Posted by Viral
Find the least value of n for which $\sum_{r=1}^n (4r-3) > 2000$

How do I do this?

$4\sum^{n}_{r=1}r-\sum^{n}_{r=1}3>2000$

$4(\frac{1}{2})n(n+1)-3n>2000$

$2n^2-n-2000>0$

Can you at least solve this inequality ?
• Oct 18th 2009, 09:47 AM
Viral
Quote:

Originally Posted by Viral
$S_n = \frac{n}{2}[2a + (n-1)d]$

Just to check, is this the formula for calculating the sum?

I'm trying to do it by using this.
• Oct 18th 2009, 05:13 PM
mr fantastic
Quote:

Originally Posted by Viral
I'm trying to do it by using this.

The left hand side of the inequality given by mathaddict, regardless of how it was derived, is exactly what you will get if you've used the formula you're using correctly and simplified correctly. So the question put to you remains: Can you solve that inequality?

(If you don't get that left hand side then you've made a mistake and will need to show all your work if additional help is to be given).
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