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The title is bad, but I was unable to think of better one, I apologize for this. I have this system: \begin{align*} a_2 + a_3 + a_4 + a_5 & = 34\\ a_2 * a_5 & = 52 \end{align*}

I have to find the $20$th member of the progression ($a_{20}$). Also I know that the progression is growing e.g. $a_1 < a_2 < a_3 \ldots$ Any ideas how to solve this task?

I've used the formula $a_n = a_1 + (n-1)d$, but I get an unrealistic equation.

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    $\begingroup$ The most straightforward, unclever way to begin is to simply use the formula $a_n=a_1+(n-1)d$, then you will have two equations in the two unknowns $a_1$ and $d$. $\endgroup$
    – anon
    Commented Jan 19, 2015 at 20:39
  • $\begingroup$ I've done that, i'm getting quadratic equation which seems unrealistic, however in the exercise is said that the progression is growing, so I guess I must have quadraci equation. $\endgroup$
    – user209047
    Commented Jan 19, 2015 at 20:43

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You can set up a system of two equations. First off, $a_2+a_5=17$ because $a_2+a_5=a_3+a_4$ So your system is $$a_2+a_5=17\\a_2\times a_5=52$$ Do you know how to solve that system?

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  • $\begingroup$ I'm getting to quadratic equation but it look unrealistic because the discriminant = 160^2 - 4*(-9)*235 $\endgroup$
    – user209047
    Commented Jan 19, 2015 at 20:41
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    $\begingroup$ The quadratic I get is $x^2-17x+52=0$ $\endgroup$ Commented Jan 19, 2015 at 20:45
  • $\begingroup$ I have mistake somewhere, i will fix it thank you. $\endgroup$
    – user209047
    Commented Jan 19, 2015 at 20:48

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