# another arithmetic series problem

• Nov 2nd 2008, 02:00 AM
pop_91
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
• Nov 2nd 2008, 02:07 AM
Moo
Hello,
Quote:

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 (Surprised)

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

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

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

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

$\displaystyle (n+1)^2>1001$

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

Does it look clear ?
• Nov 2nd 2008, 02:18 AM
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
• Nov 2nd 2008, 02:25 AM
Moo
Quote:

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 !

$\displaystyle 2n+4=2(n+2)$
so $\displaystyle \frac n2 (2n+4)=\frac{n \cdot {\color{red}2} (n+2)}{\color{red}2}=n(n+2)$
• Nov 2nd 2008, 02:38 AM
pop_91
oooh! thanks i get it now x