# Arithmetic Progression [closed]

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

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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

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• 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. – kjo May 28 '16 at 17:39

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