Question

sa se afle s30 al progresiei aritmetice an in care a1+a7=42 a10-a3=21 in acolada cele doua

Answer 1

$\displaystyle \left \{ {{a_1+a_7=42} \atop {a_{10}-a_3=21}} \right. \Rightarrow \left \{ {{a_1+a_{7-1}+r=42} \atop {(a_{10-1}+r)-(a_{3-1}+r)=21}} \right. \Rightarrow \\ \Rightarrow \left \{ {{a_1+a_6+r=42} \atop {(a_9+r)-(a_2+r)=21}} \right. \Rightarrow \left \{ {{a_1+a_1+6r=42} \atop {(a_1+9r)-(a_1+2r)=21}} \right. \Rightarrow \\ \Rightarrow \left \{ {{2a_1+6r=42} \atop {a_1+9r-a_1-2r=21}} \right. \Rightarrow \left \{ {{2a_1+6r=42} \atop {~~~~~~~7r=21}} \right.$
$\displaystyle 7r=21 \Rightarrow r= \frac{21}{7} \Rightarrow r=3 \\ 2a_1+6r=42 \Rightarrow 2a_1+6 \cdot 3=42 \Rightarrow 2a_1+18=42 \Rightarrow 2a_1=42-18 \Rightarrow \\ \Rightarrow 2a_1=24 \Rightarrow a_1= \frac{24}{2} \Rightarrow a_1=12 \\ S_{30}= \frac{2 \cdot 12+(30-1) \cdot 3}{2} \cdot 30 \\ S_{30}=(24+29 \cdot 3) \cdot 15 \\ S_{30}=(24+87) \cdot 15 \\ S_{30}=111 \cdot 15 \\ S_{30}=1665$