# Math Help - another arithmetic series problem

1. ## another arithmetic series problem

an artihmetic series has a first term of 3 and a common difference of 2

determine the number of terms needed for te sum to exceed 1000

I think I'm missing something blindingly obvious but i can't seem to finish it, but at the moment i'm this far...

1000< n/2(2n+4)

now I'm stuck....

any help is greatly recieved

2. Hello,
Originally Posted by pop_91
an artihmetic series has a first term of 3 and a common difference of 2

determine the number of terms needed for te sum to exceed 1000

I think I'm missing something blindingly obvious but i can't seem to finish it, but at the moment i'm this far...

1000< n/2(2n+4)

now I'm stuck....

any help is greatly recieved
Perfect so far

Well now :
$\frac n2 (2n+4)=n(n+2)$

So you want to solve $n(n+2)>1000$
$n^2+2n>1000$

$n^2+2n+1-1>1000$

$(n+1)^2-1>1000$

$(n+1)^2>1001$

In fact, if you want to find the minimum value of n, solve $(n+1)^2=1001$ and then round it up by superior value.

Does it look clear ?

3. yes i think i understand just one question...

do you not have to divide 1000 by 2 because you have to do the same to both sides.... i'm not sure

4. Originally Posted by pop_91
yes i think i understand just one question...

do you not have to divide 1000 by 2 because you have to do the same to both sides.... i'm not sure
Oh !

$2n+4=2(n+2)$
so $\frac n2 (2n+4)=\frac{n \cdot {\color{red}2} (n+2)}{\color{red}2}=n(n+2)$

5. oooh! thanks i get it now x