Question

Ajutati-ma va rog. Matematica M2

Answer 1

$\text{Rezolvarea pentru 25:}\\ \\\left\{ \begin{array}{ll} a_1+2a_3=4 \\ a_2+2a_4=12 \end{array} \right ,~~\text{Se cere: }19\cdot q\cdot a_1, \text{ adica: }19a_2\\ \\ \text{Scriem tot in functie de }a_2.\\ \\ a_2 = a_1\cdot q \Rightarrow a_1 = \dfrac{a_2}{q}\\ a_3 = a_2\cdot q \\ a_4 = a_3\cdot q = a_2\cdot q\cdot q = a_2\cdot q^2$

$\Rightarrow \left\{ \begin{array}{ll} \dfrac{a_2}{q}+2a_2\cdot q=4\Big|\cdot q \\ a_2+2a_2 \cdot q^2 =12 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} a_2+2a_2\cdot q^2=4\cdot q \\ a_2+2a_2 \cdot q^2 =12 \end{array} \right \Rightarrow \\ \\ \Rightarrow 4\cdot q = 12 \Rightarrow q = 3 \\ \\ \text{Inlocuim q in ultimul sistem de ecuatii:} \\ \\ \Rightarrow \left\{ \begin{array}{ll} a_2+2a_2\cdot 3^2=4\cdot 3 \\ a_2+2a_2 \cdot 3^2 =12 \end{array} \right \Rightarrow a_2+2a_2\cdot 3^2 = 12 \Rightarrow$

$\Rightarrow a_2 + 2a_2\cdot 9 = 12 \Rightarrow a_2+18a_2 = 12 \Rightarrow 19a_2 =12 \Rightarrow\\ \\ \Rightarrow \boxed{19\cdot q\cdot a_1 = 12}$