Question

Să se determine primul termen și rația progresiei aritmetice (an) dacă:
a) a4=10, a7=19

b) a3= -4, S15= -285

c) {a2+a5-a8=10
{a1=a6=17

d) {a4+a8=30
{10a1-4a7= -45

e) {a2+a6+a9=45
{a3+a7+a10=54

f) {S3=12
{S6=51

g) {S2-S4+a2=14
{S3+a3=17

Answer 1

[tex]\displaystyle a).a_4=10,~a_7=19 \\ a_4=10 \Rightarrow a_{4-1}+r=10 \Rightarrow a_3+r=10 \Rightarrow a_1+3r=10 \Rightarrow \\ \Rightarrow a_1=10-3r \\ a_7=19 \Rightarrow a_{7-1}+r=19 \Rightarrow a_6+r=19 \Rightarrow a_1+6r=19 \Rightarrow \\ \Rightarrow 10-3r+6r=19 \Rightarrow -3r+6r=19-10 \Rightarrow 3r=9 \Rightarrow r= \frac{9}{3} \Rightarrow \\ \Rightarrow \boxed{r=3} \\ a_1=10-3r \Rightarrow a_1=10-3 \cdot 3 \Rightarrow a_1=10-9 \Rightarrow \boxed{a_1=1} [/tex]
[tex]\displaystyle b).a_3=-4,~S_{15}=-285\\a_3=-4 \Rightarrow a_{3-1}+r=-4 \Rightarrow a_2+r=-4 \Rightarrow a_1+2r=-4 \Rightarrow \\ \Rightarrow a_1=-4-2r \\ S_{15}=-285 \Rightarrow \frac{2(-4-2r)+(15-1) \cdot r}{2} \cdot 15 =-285 \Rightarrow \\ \Rightarrow \frac{-8-4r+14r}{2}\cdot 15=-285 \Rightarrow \frac{-8+10r}{2} \cdot15=-285 \Rightarrow \\ \Rightarrow -120+150r=2 \cdot (-285) \Rightarrow -120+150r=-570\Rightarrow [/tex]
[tex]\displaystyle \Rightarrow150r=-570+120 \Rightarrow 150r=-450 \Rightarrow r= - \frac{450}{150} \Rightarrow \boxed{r=-3} \\ a_1=-4-2r \Rightarrow a_1=-4-2 \cdot (-3) \Rightarrow a_1=-4+6 \Rightarrow \boxed{a_1=2} [/tex]
[tex]\displaystyle c). \left \{ {{a_2+a_5-a_8=10} \atop {a_1+a_6=17}} \right. \Rightarrow \left \{ {{a_{2-1}+r+a_{5-1}+r-(a_{8-1}+r)=10} \atop {a_1+a_{6-1}+r=17}} \right. \Rightarrow \\ \Rightarrow \left \{ {{a_1+r+a_4+r-(a_7+r)=10} \atop {a_1+a_5+r=17}} \right. \Rightarrow \\ \Rightarrow \left \{ {{a_1+r+a_1+4r-(a_1+7r)=10} \atop {a_1+a_1+5r=17}} \right. \Rightarrow [/tex]
[tex]\displaystyle \Rightarrow \left \{ {{a_1+r+a_1+4r-a_1-7r=10} \atop {2a_1+5r=17}} \right. \Rightarrow \left \{ {{a_1-2r=10/\cdot (-2)} \atop {2a_1+5r=17}} \right. \Rightarrow \\ \Rightarrow \left \{ {{-2a_1+4r=-20} \atop {2a_1+5r=17}} \right. \\ ~~~~--------- \\ ~~~~~~~~~~/ ~~~~~~9r~= -3 \Rightarrow r=- \frac{3}{9} \Rightarrow \boxed{r=- \frac{1}{3}} [/tex]
[tex]\displaystyle a_1-2r=10 \Rightarrow a_1-2 \cdot \left(- \frac{1}{3} \right)=10 \Rightarrow a_1+ \frac{2}{3} =10 \Rightarrow 3a_1+2=30 \Rightarrow \\ \Rightarrow 3a_1=30-2 \Rightarrow 3a_1=28 \Rightarrow \boxed{a_1= \frac{28}{3} }[/tex]
[tex]\displaystyle d). \left \{ {{a_4+a_8=30} \atop {10a_1-4a_7=-45}} \right. \Rightarrow \left \{ {{a_{4-1}+r+a_{8-1}+r=30} \atop {10a_1-4(a_{7-1}+r)=-45}} \right. \Rightarrow \\ \Rightarrow \left \{ {{a_3+r+a_7+r=30} \atop {10a_1-4(a_6+r)=-45 }} \right. \Rightarrow \left \{ {{a_1+3r+a_1+7r=30} \atop {10a_1-4(a_1+6r)=-45}} \right. \Rightarrow \\ \Rightarrow \left \{ {{2a_1+10r=30} \atop {10a_1-4a_1-24r=-45}} \right. \Rightarrow \left \{ {{2a_1+10r=30/\cdot(-3)} \atop {6a_1-24r=-45}} \right. \Rightarrow[/tex]
[tex]\displaystyle \Rightarrow \left \{ {{-6a_1-30r=-90} \atop {6a_1-24r=-45}} \right. \\ ~~~~---------- \\ ~~~~~~~~~/ ~~~-54r=-135 \Rightarrow r= \frac{135}{54} \Rightarrow \boxed{r= \frac{5}{2}} \\ 2a_1+10r=30 \Rightarrow 2a_1+10 \cdot \frac{5}{2} =30 \Rightarrow 2a_1+ \frac{50}{2} =30 \Rightarrow \\ \Rightarrow 2a_1+25=30 \Rightarrow 2a_1=30-25 \Rightarrow 2a_1=5 \Rightarrow \boxed{a_1= \frac{5}{2}} [/tex]
[tex]\displaystyle e). \left \{ {{a_2+a_6+a_9=45} \atop {a_3+a_7+a_{10}=54}} \right. \Rightarrow \left \{ {{a_{2-1}+r+a_{6-1}+r+a_{9-1}+r=45} \atop {a_{3-1}+r+a_{7-1}+r+a_{10-1}+r=54}} \right. \Rightarrow \\ \Rightarrow \left \{ {{a_1+r+a_5+r+a_8+r=45} \atop {a_2+r+a_6+r+a_9+r=54}} \right. \Rightarrow \\ \Rightarrow \left \{ {{a_1+r+a_1+5r+a_1+8r=45} \atop {a_1+2r+a_1+6r+a_1+9r=54}} \right. \Rightarrow \left \{ {{3a_1+14r=45/ \cdot (-1)} \atop {3a_1+17r=54}} \right. \Rightarrow[/tex]
[tex]\displaystyle \Rightarrow \left \{ {{-3a_1-14r=-45} \atop {3a_1+17r=54}} \right. \\ ~~~~~~~~~~~/ ~~~~~~3r=9 \Rightarrow r= \frac{9}{3} \Rightarrow \boxed{r=3} \\ 3a_1+14r=45 \Rightarrow 3a_1+14 \cdot 3=45 \Rightarrow 3a_1+42=45 \Rightarrow \\ \Rightarrow 3a_1=45-42 \Rightarrow 3a_1=3 \Rightarrow a_1= \frac{3}{3} \Rightarrow \boxed{a_1=1} [/tex]