# Arithmetic series question

• Jan 31st 2009, 11:35 PM
milan
Arithmetic series question
1. In an arithmetic series, three consecutive terms have a sum of -9 and a product of 48. Find the possible values of these terms.

I really cant figure it out. I tries using the Sum of series formula and arithmetic series formula, but no results.
• Jan 31st 2009, 11:42 PM
Moo
Hello,
Quote:

Originally Posted by milan
1. In an arithmetic series, three consecutive terms have a sum of -9 and a product of 48. Find the possible values of these terms.

I really cant figure it out. I tries using the Sum of series formula and arithmetic series formula, but no results.

Let k be the constant progression of the series.
Let $\displaystyle a,b,c$ the three consecutive terms.
We have $\displaystyle a+b+c=-9$, $\displaystyle abc=48$
But we also know that $\displaystyle b=a+k$ and $\displaystyle c=a+2k$

Does this help ?
• Jan 31st 2009, 11:59 PM
milan
I know all that, but i dont know how to find the answer using it:P, thanx anyways though.
• Feb 1st 2009, 12:08 AM
Moo
Substituting $\displaystyle b=a+k$ and $\displaystyle c=a+2k$ in $\displaystyle a+b+c=-9$, you have :

$\displaystyle 3a+3k=-9 \Rightarrow a+k=-3$

Now substituting in $\displaystyle abc=48$, you get :
$\displaystyle a(a+k)(a+2k)=48$
We know that a+k=-3.
Hence $\displaystyle -a(a+2k)=16$ and furthermore, $\displaystyle a=-3-k$, so :
$\displaystyle -(-3-k)(-3-k+2k)=16$
$\displaystyle (3+k)(-3+k)=16$
$\displaystyle k^2-9=16$

So $\displaystyle k=\pm 5$

If $\displaystyle k=5$, then $\displaystyle a=-8$ and $\displaystyle b=\dots~,~c=\dots$

If $\displaystyle k=-5$, then ......................

Does it look better ?