Question

1.Să se determine rația și suma primilor n termeni pentru progresia aritmetica (An) daca :
a) a1 = 5, a26= 105, n=26;
b) a1 = -10, an=-20, n=6.

2.Sa se determine primul termen și rația progresiei aritmetice (An) daca :
a) {a2 + a5 - a8= 10
{a1+a6=17

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

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

d) {S3=12
{S6=51

e) {5a3 - 4a8 = 179
{2S5 - 3S2 = 4

Vă rog , ajutați-mă! Dau coroana.

Answer 1

$1a).\displaystyle a_1=5,~a_{26}=105,~n=26 \\ a_{26}=105 \Rightarrow a_{26-1}+r=105 \Rightarrow a_{25}+r=105 \Rightarrow a_1+25r=105 \Rightarrow \\ \Rightarrow 5+25r=105 \Rightarrow 25r=105-5 \Rightarrow 25r=100 \Rightarrow r= \frac{100}{25} \Rightarrow \boxed{r=4} \\ S_{26}= \frac{2 \cdot 5+(26-1) \cdot 4}{\not2} \cdot \not26 \\ S_{26}=(10+25 \cdot 4) \cdot 13 \\ S_{26}=(10+100) \cdot 13 \\ S_{26}=110 \cdot 13 \\ \boxed{S_{26}=1430}$
$\displaystyle b).a_1=-10,~a_n=-20,~n=6 \\ a_n=a_1+(n-1) \cdot r \Rightarrow -20=-10+(6-1) \cdot r \Rightarrow -20=-10+5r \Rightarrow \\ \Rightarrow 5r=-20+10 \Rightarrow 5r=-10 \Rightarrow r=- \frac{10}{5} \Rightarrow \boxed{r=-2} \\ S_{6}= \frac{-10-20}{\not2} \cdot \not6 \\ S_6=-30 \cdot 3 \\ \boxed{S_6=-90}$
$\displaystyle 2a). \left \{ {{a_2+a_5-a_8=10} \atop {a_1+a_6=17}} \right. \Rightarrow \left \{ {{a_{2-1}+r+a_{5-1}+r-(a_{8-1}+r)=10} \atop {a_1+a_{6-1}+r=17}} \right. \Rightarrow \\ \Rightarrow \left \{ {{a_1+r+a_4+r-(a_7+r)=10} \atop {a_1+a_5+r=17}} \right. \Rightarrow \\ \Rightarrow \left \{ {{a_1+r+a_1+4r-(a_1+7r)=10} \atop {a_1+a_1+5r=17}} \right. \Rightarrow$
$\displaystyle \Rightarrow \left \{ {{a_1+r+a_1+4r-a_1-7r=10} \atop {2a_1+5r=17}} \right. \Rightarrow \left \{ {{a_1-2r=10/\cdot (-2)} \atop {2a_1+5r=17}} \right. \Rightarrow \\ \Rightarrow \left \{ {{-2a_1+4r=-20} \atop {2a_1+5r=17}} \right. \\ ~~~~--------- \\ ~~~~~~~~~~/ ~~~~~~9r~= -3 \Rightarrow r=- \frac{3}{9} \Rightarrow \boxed{r=- \frac{1}{3}}$
$\displaystyle a_1-2r=10 \Rightarrow a_1-2 \cdot \left(- \frac{1}{3} \right)=10 \Rightarrow a_1+ \frac{2}{3} =10 \Rightarrow 3a_1+2=30 \Rightarrow \\ \Rightarrow 3a_1=30-2 \Rightarrow 3a_1=28 \Rightarrow \boxed{a_1= \frac{28}{3} }$
$\displaystyle b). \left \{ {{a_4+a_8=30} \atop {10a_1-4a_7=-45}} \right. \Rightarrow \left \{ {{a_{4-1}+r+a_{8-1}+r=30} \atop {10a_1-4(a_{7-1}+r)=-45}} \right. \Rightarrow \\ \Rightarrow \left \{ {{a_3+r+a_7+r=30} \atop {10a_1-4(a_6+r)=-45 }} \right. \Rightarrow \left \{ {{a_1+3r+a_1+7r=30} \atop {10a_1-4(a_1+6r)=-45}} \right. \Rightarrow \\ \Rightarrow \left \{ {{2a_1+10r=30} \atop {10a_1-4a_1-24r=-45}} \right. \Rightarrow \left \{ {{2a_1+10r=30/\cdot(-3)} \atop {6a_1-24r=-45}} \right. \Rightarrow$
$\displaystyle \Rightarrow \left \{ {{-6a_1-30r=-90} \atop {6a_1-24r=-45}} \right. \\ ~~~~---------- \\ ~~~~~~~~~/ ~~~-54r=-135 \Rightarrow r= \frac{135}{54} \Rightarrow \boxed{r= \frac{5}{2}} \\ 2a_1+10r=30 \Rightarrow 2a_1+10 \cdot \frac{5}{2} =30 \Rightarrow 2a_1+ \frac{50}{2} =30 \Rightarrow \\ \Rightarrow 2a_1+25=30 \Rightarrow 2a_1=30-25 \Rightarrow 2a_1=5 \Rightarrow \boxed{a_1= \frac{5}{2}}$
$\displaystyle c). \left \{ {{a_2+a_6+a_9=45} \atop {a_3+a_7+a_{10}=54}} \right. \Rightarrow \left \{ {{a_{2-1}+r+a_{6-1}+r+a_{9-1}+r=45} \atop {a_{3-1}+r+a_{7-1}+r+a_{10-1}+r=54}} \right. \Rightarrow \\ \Rightarrow \left \{ {{a_1+r+a_5+r+a_8+r=45} \atop {a_2+r+a_6+r+a_9+r=54}} \right. \Rightarrow \\ \Rightarrow \left \{ {{a_1+r+a_1+5r+a_1+8r=45} \atop {a_1+2r+a_1+6r+a_1+9r=54}} \right. \Rightarrow \left \{ {{3a_1+14r=45/ \cdot (-1)} \atop {3a_1+17r=54}} \right. \Rightarrow$
$\displaystyle \Rightarrow \left \{ {{-3a_1-14r=-45} \atop {3a_1+17r=54}} \right. \\ ~~~~~~~~~~~/ ~~~~~~3r=9 \Rightarrow r= \frac{9}{3} \Rightarrow \boxed{r=3} \\ 3a_1+14r=45 \Rightarrow 3a_1+14 \cdot 3=45 \Rightarrow 3a_1+42=45 \Rightarrow \\ \Rightarrow 3a_1=45-42 \Rightarrow 3a_1=3 \Rightarrow a_1= \frac{3}{3} \Rightarrow \boxed{a_1=1}$