Question

să se afle primul termen și rația unei progresii aritmetice dacă

2a4 -3a2 + a12 = 28
a5 × a7 = 45

Answer 1

 

$\displaystyle\\\begin {cases}2a_4 -3a_2 + a_{12} = 28\\a_5 \times a_7 = 45\end {cases}\\\\\\\begin {cases}2(a_2+2r) -3a_2 + (a_2+10r) = 28\\(a_2+3r) \times (a_2+5r) = 45\end {cases}\\\\\\\begin {cases}2a_2+4r -3a_2 + a_2+10r = 28\\(a_2+3r) \times (a_2+5r) = 45\end {cases}\\\\\\\begin {cases}2a_2-3a_2+ a_2+4r + 10r = 28\\(a_2+3r) \times (a_2+5r) = 45\end {cases}\\\\\\\begin {cases}14r = 28 \\(a_2+3r) \times (a_2+5r) = 45\end {cases}\\\\r=\frac{28}{14}\\\\\boxed{r=2}$

$\displaystyle\\(a_2+3r) \times (a_2+5r) = 45\\\\(a_2+3\times2) \times (a_2+5\times2) = 45\\\\(a_2+6) \times (a_2+10) = 45\\\\a_2^2+16a_2+60=45\\\\a_2^2+16a_2+15=0\\\\\text{Vom obtine 2 solutii pentru } a_2\\\\a_2=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\\\\=\frac{-16\pm\sqrt{256-60}}{2}=\frac{-16\pm\sqrt{196}}{2}=\frac{-16\pm14}{2}=-8\pm7$

$\displaystyle\\\bf Solutia~1:\\a_2=-8-7=-15~~si~~r=2\\a_1=a2-r=-15-2=-17\\\implies~~\boxed{\bf a_1=-17~~si~~r=2}\\\\\\Solutia~2:\\a_2=-8+7=-1~~si~~r=2\\a_1=a2-r=-1-2=-3\\\implies~~\boxed{\bf a_1=-3~~si~~r=2}$