-4
$\begingroup$

How can I resolve these issues? I'm from Brazil and I don't have many things related to the subject here

**

  1. In an arithmetic progression of 6 Terms, the reason is 5. The product of the 1° with the latter is 12500. Determine the value of 3° Term Ps. Consider the Arithmetic Progression of positive terms

**

$\endgroup$

closed as off-topic by user9660, m_goldberg, C. E., kjo, WReach May 28 '16 at 17:52

  • The question does not concern the technical computing software Mathematica by Wolfram Research. Please see the help center to find out about the topics that can be asked here.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ I'm voting to close this question as off-topic because the question is almost completely incoherent, and mentions nothing whatsoever directly related to Mathematica. $\endgroup$ – kjo May 28 '16 at 17:39
2
$\begingroup$

If I understand your question, there is no need of Mathematica to solve your problem

for $x_0$ given, we know that $y_{n+1} = y_{n} + 5$. We know also that $y_0 y_1 = 12500$. That is to say that $y_0 y_1 = y_0 (y_0 + 5) = y_0^2 + 5 y_0 = 12500$. You can ask Mathematica to solve this second order equation to obtain $x_0 = 109.331$ --- there is also a negative root but we can eliminate it. Now we can evaluate $x_2 = 5 + y_1 = 5 + (5+y_0) = 10 + y_0 = 119.331$.

Obviously $x_2$ is the third term of the progression. Hope to have been helping !

$\endgroup$
  • $\begingroup$ Probably (since this results in nice integers) the OP meant $y_0 y_5=12500$. Then one gets $y_0=100$. $\endgroup$ – Bruno Le Floch Jun 1 '16 at 16:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.