I have been trying to solve this arithmetical sequence but I can't find a1;a2;a3 if:
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: $\displaystyle a_n = a_1 + \left( {n - 1} \right)v$
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).
So we have then
$\displaystyle \left\{ \begin{array}{l}
a_1 + a_2 + a_3 = - 12 \\
a_1 \cdot a_2 \cdot a_3 = 80 \\
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a_1 + \left( {a_1 + v} \right) + \left( {a_1 + 2v} \right) = - 12 \\
a_1 \cdot \left( {a_1 + v} \right) \cdot \left( {a_1 + 2v} \right) = 80 \\
\end{array} \right$
$\displaystyle \left\{ \begin{array}{l}
3a_1 + 3v = - 12 \\
a_1 ^3 + 3a_1 ^2 v + 2a_1 v^2 = 80 \\
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a_1 + v = - 4 \\
a_1 ^3 + 3a_1 ^2 v + 2a_1 v^2 = 80 \\
\end{array} \right$
From the first equation we have that v = -4 -a_1, substitute this in the second equation, the cubic term will disappear:
$\displaystyle a_1 ^3 + 3a_1 ^2 \left( { - 4 - a_1 } \right) + 2a_1 \left( { - 4 - a_1 } \right)^2 = 80 \Leftrightarrow 4a^2 + 32a = 80$
Solving this should be doable
Countinuing where TD! left of, we have that,
$\displaystyle a_2=a_1+k$
$\displaystyle a_3=a_1+2k$
Where $\displaystyle k$ is the common difference.
Thus,
$\displaystyle a_1+a_2+a_3=-12$
Becomes,
$\displaystyle 3a_1+3k=-12$
Thus,
$\displaystyle a_1+k=-4$ (1)
Now for the second equation,
$\displaystyle a_1(a_1+k)(a_1+2k)=80$
Thus,
$\displaystyle a_1(-4)(a_1+2k)=80$ by equation (1)
Thus,
$\displaystyle a_1(a_1+2k)=-20$
Simply (2) as,
$\displaystyle a_1(a_1+k+k)=-20$
Thus,
$\displaystyle a_1(k-4)=-20$
Thus,
$\displaystyle a_1k-4a_1=-20$ (2)
Now you have the two equations, solve them,
substitute for $\displaystyle a_1$ from (1),
$\displaystyle (-4-k)k-4(-4-k)=-20$
Thus,
$\displaystyle -4k-k^2+16+4k=-20$
Simplify,
$\displaystyle k^2=36$
Thus,
$\displaystyle (k,a_1)=(6,-10)\mbox{ or }(-6,-14)$
Now just check which one works.
Q.E.D.
There's a slight mistake here, for k = -6 we have that a_1 = 2.Originally Posted by ThePerfectHacker
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.Originally Posted by TD!
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.
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).
That's why I asked if it was a_3 or a_5. From what I understand now, it was actually a_5 used in the equation but you were asked to find a_1,a_2 and a_3?
In that case, just repeat the exact same method but use a_5 instead of a_3, a_5 can be converted to a_1 + 4v.
You have the entire method now, try it