# Arithmetical sequence

• Jan 27th 2006, 09:25 AM
totalnewbie
Arithmetical sequence
I have been trying to solve this arithmetical sequence but I can't find a1;a2;a3 if:
• Jan 27th 2006, 10:25 AM
TD!
Is it a_3 or a_5?

Anway, use the basic property of an arithmetic sequence to convert everything in terms of a_1 and the difference of the sequence, I'll call that v.

We then have: $a_n = a_1 + \left( {n - 1} \right)v$

Use this on all terms except a_1 of course and then you have a system of two equations in two unknowns (a_1 and v).
• Jan 27th 2006, 11:21 AM
totalnewbie
It is a_3.
I used the formula what you mentioned and I substituted v into the second operation. I got cube equation -7a_1^3-6a_1^2+288a_1=1000
I am not able to find the a_1 from that cube equation.
• Jan 27th 2006, 11:26 AM
TD!
So we have then

$\left\{ \begin{array}{l}
a_1 + a_2 + a_3 = - 12 \\
a_1 \cdot a_2 \cdot a_3 = 80 \\
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a_1 + \left( {a_1 + v} \right) + \left( {a_1 + 2v} \right) = - 12 \\
a_1 \cdot \left( {a_1 + v} \right) \cdot \left( {a_1 + 2v} \right) = 80 \\
\end{array} \right$

$\left\{ \begin{array}{l}
3a_1 + 3v = - 12 \\
a_1 ^3 + 3a_1 ^2 v + 2a_1 v^2 = 80 \\
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a_1 + v = - 4 \\
a_1 ^3 + 3a_1 ^2 v + 2a_1 v^2 = 80 \\
\end{array} \right$

From the first equation we have that v = -4 -a_1, substitute this in the second equation, the cubic term will disappear:

$a_1 ^3 + 3a_1 ^2 \left( { - 4 - a_1 } \right) + 2a_1 \left( { - 4 - a_1 } \right)^2 = 80 \Leftrightarrow 4a^2 + 32a = 80$

Solving this should be doable :)
• Jan 27th 2006, 11:31 AM
ThePerfectHacker
Countinuing where TD! left of, we have that,
$a_2=a_1+k$
$a_3=a_1+2k$
Where $k$ is the common difference.
Thus,
$a_1+a_2+a_3=-12$
Becomes,
$3a_1+3k=-12$
Thus,
$a_1+k=-4$ (1)
Now for the second equation,
$a_1(a_1+k)(a_1+2k)=80$
Thus,
$a_1(-4)(a_1+2k)=80$ by equation (1)
Thus,
$a_1(a_1+2k)=-20$
Simply (2) as,
$a_1(a_1+k+k)=-20$
Thus,
$a_1(k-4)=-20$
Thus,
$a_1k-4a_1=-20$ (2)
Now you have the two equations, solve them,
substitute for $a_1$ from (1),
$(-4-k)k-4(-4-k)=-20$
Thus,
$-4k-k^2+16+4k=-20$
Simplify,
$k^2=36$
Thus,
$(k,a_1)=(6,-10)\mbox{ or }(-6,-14)$
Now just check which one works.
Q.E.D.
• Jan 27th 2006, 11:39 AM
TD!
Quote:

Originally Posted by ThePerfectHacker
Thus,
$(k,a_1)=(6,-10)\mbox{ or }(-6,-14)$
Now just check which one works.
Q.E.D.

There's a slight mistake here, for k = -6 we have that a_1 = 2.

Furthermore, there is not reason to check which one 'works' since there's no physical interpretation or whatever which has to be checked. All we did was solving the system of equations algebraiclly so both of our solutions are fine, i.e. meet the conditions which were given.
• Jan 27th 2006, 11:46 AM
ThePerfectHacker
Quote:

Originally Posted by TD!
There's a slight mistake here, for k = -6 we have that a_1 = 2.

Furthermore, there is not reason to check which one 'works' since there's no physical interpretation or whatever which has to be checked. All we did was solving the system of equations algebraiclly so both of our solutions are fine, i.e. meet the conditions which were given.

I erred again, but it is the method not the answer that is important.
You statement about the checking the two solutions; I know the two solutions work for the algebraic equations but not necessarily for the conditions given for the problem thus we must check.
• Jan 27th 2006, 11:48 AM
TD!
Well the problem was that knowing we were dealing with an arithmetic sequence, "find a1;a2;a3 if: (the two equations)".
Both solutions are fine, there is no reason to discard one since there's nothing left to check! Unless, of course, totalnewbie didn't give the entire problem (but I think he did or at least I assumed so).
• Jan 27th 2006, 12:19 PM
totalnewbie
It seems to me that my math book contains a mistake.
There is a_1+a_2+a_5=-12