Question

Determinati in fiecare caz,primul termen si ratia progresei aritmetice $a_{n}$:
a) $a_{1}$ + $a_{4}$ = 14 ; $a_{2}$ + $a_{5}$ = 18
b) $a_{2}$ + $a_{4}$ = 20 ; $a_{3}$ + $a_{5}$ = 26
c) $a_{3}$ - $a_{2}$ = 3 ; $a_{4}$ + $a_{6}$ = 28

Answer 1

$\displaystyle a) \left \{ {{a_1+a_4=14} \atop {a_2+a_5=18}} \right. \Rightarrow \left \{ {{a_1+a_{4-1}+r=14} \atop {a_{2-1}+r+a_{5-1}+r=18}} \right. \Rightarrow \\ \\ \Rightarrow \left \{ {{a_1+a_3+r=14} \atop {a_1+r+a_4+r=18}} \right. \Rightarrow \left \{ {{a_1+a_1+3r=14} \atop {a_1+r+a_1+4r=18}} \right. \Rightarrow$

$\displaystyle \Rightarrow \left \{ {{2a_1+3r=14|\cdot(-1)} \atop 2a_1+5r=18}} \right. \Rightarrow \left \{ {{-2a_1-3r=-14} \atop {2a_1+5r=18}} \right. \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~----------\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~/~~~~~~~2r=4 \Rightarrow r= \frac{4}{2} \Rightarrow \boxed{r=2} \\ \\ 2a_1+3 \cdot 2=14 \Rightarrow 2a_1+6=14 \Rightarrow2a_1=14-6 \Rightarrow 2a_1=8 \Rightarrow \\ \\ \Rightarrow a_1= \frac{8}{2} \Rightarrow \boxed{a_1=4}$

$\displaystyle b) \left \{ {{a_2+a_4=20} \atop {a_3+a_5=26}} \right. \Rightarrow \left \{ {{a_{2-1}+r+a_{4-1}+r=20} \atop {a_{3-1}+r+a_{5-1}+r=26}} \right. \Rightarrow \\ \\ \left \{ {{a_1+r+a_3+r=20} \atop {a_2+r+a_4+r=26}} \right. \Rightarrow \left \{ {{a_1+r+a_1+3r=20} \atop {a_1+2r+a_1+4r=26}}\right. \Rightarrow$

$\displaystyle \Rightarrow \left \{ {{2a_1+4r=20|\cdot(-1)} \atop {2a_1+6r=26}} \right. \Rightarrow \left \{ {{-2a_1-4r=-20} \atop {2a_1+6r=26}} \right. \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~----------\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~/~~~~~2r=6 \Rightarrow r= \frac{6}{2} \Rightarrow \boxed{r=3}\\ \\ 2a_1 +4 \cdot3=20 \Rightarrow 2a_1+12=20 \Rightarrow 2a_1=20-12 \Rightarrow 2a_1=8 \Rightarrow \\ \\ a_1= \frac{8}{2} \Rightarrow \boxed{a_1=4}$

$\displaystyle c)\left \{ {{a_3-a_2=3} \atop {a_4+a_6=28}} \right. \Rightarrow \left \{ {{a_{3-1}+r-(a_{2-1}+r)=3} \atop {a_{4-1}+r+a_{6-1}+r=28}} \right. \Rightarrow \\ \\ \Rightarrow \left \{ {{a_2+r-(a_1+r)=3} \atop {a_3+r+a_5+r=28}} \right. \Rightarrow \left \{ {{a_1+2r-a_1-r=3} \atop {a_1+3r+a_1+5r=28}} \right. \Rightarrow \\ \\ \Rightarrow \left \{ {{\boxed{r=3}} \atop {2a_1+8r=28}} \right.$

$\displaystyle 2a_1+8 \cdot 3=28 \Rightarrow 2a_1+24=28 \Rightarrow 2a_1=28-24 \Rightarrow 2a_1=4 \Rightarrow \\ \\ \Rightarrow a_1= \frac{4}{2} \Rightarrow \boxed{a_1=2}$