# Thread: Find the least value of n in an arithmetic sequence

1. Ok, I've used that formula but don't know how to solve for n.

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

How do I do this?
$\displaystyle 4\sum^{n}_{r=1}r-\sum^{n}_{r=1}3>2000$

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

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

Can you at least solve this inequality ?

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

Just to check, is this the formula for calculating the sum?
I'm trying to do it by using this.

4. 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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# formula of least value of n>2000

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