Arithmetic series question

**milan** · January 31st 2009, 11:35 PM

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.

**Moo** · January 31st 2009, 11:42 PM

Hello,

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 $a,b,c$ the three consecutive terms.
We have $a+b+c=-9$ , $abc=48$
But we also know that $b=a+k$ and $c=a+2k$

Does this help ?

**milan** · January 31st 2009, 11:59 PM

I know all that, but i dont know how to find the answer using it:P, thanx anyways though.

**Moo** · February 1st 2009, 12:08 AM

Substituting $b=a+k$ and $c=a+2k$ in $a+b+c=-9$ , you have :

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

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

So $k=\pm 5$

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

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

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