Question

Se da o progresie aritmetică a10+a5=12 și a5-a8=13 Calculați ratia și a1=?

Answer 1

$\[ \left\{ \begin{array}{Il} a_{10} + a_5 = 12 && a_5 - a_8 = 13 \ \end{array} \Rightarrow \[ \left\{ \begin{array}{Il} a_1+9r + a_1+4r = 12 && a_1+4r - (a_1+7r) = 13 \ \end{array} \Rightarrow \right. \]$

$\[ \left\{ \begin{array}{Il} 2a_1+13r = 12 && a_1+4r - a_1-7r = 13 \ \end{array} \Rightarrow \[ \left\{ \begin{array}{Il} 2a_1+13r = 12 &&-3r = 13 \ \end{array} \Rightarrow$

$\[ \left\{ \begin{array}{Il} 2a_1+13r = 12 && r = -\dfrac{13}{3} \ \end{array} \Rightarrow \[ \left\{ \begin{array}{Il} 2a_1+13\cdot \Big(-\dfrac{13}{3}\Big) = 12 && r = -\dfrac{13}{3} \ \end{array} \Rightarrow$

$\[ \left\{ \begin{array}{Il} 2a_1 =\quad\quad \dfrac{12\cdot 3+13\cdot 13}{3\cdot 2} && r = -\dfrac{13}{3} \ \end{array} \Rightarrow \boxed{ \[ \left\{ \begin{array}{Il} a_1 = \dfrac{205}{6} && \quad \quad r = -\dfrac{13}{3} \ \end{array} }$