Question

Să se determine suma primilor n termeni al progresiei aritmetice an dacă:
a) a1= -1, r=3, n=12
b) a1=100, r=2, n=60
c) a7=17, a2=2, n=50
d) a3= -12, a5=36, n=20

Answer 1

$\displaystyle a).a_1=-1,~r=3,~n=12 ~~~~~~~~~~~~~~~~~~~\boxed{S_n= \frac{2a_1+(n-1) \cdot r}{2} \cdot n} \\ S_{12}= \frac{2 \cdot (-1)+(12-1) \cdot 3}{\not2} \cdot \not12 \\ S_{12}=(-2+11 \cdot 3) \cdot 6 \\ S_{12}=(-2+33) \cdot 6 \\ S_{12}=31 \cdot 6 \\ S_{12}=186$
$\displaystyle b).a_1=100,~r=2,~n=60 \\ S_{60}= \frac{2 \cdot 100+(60-1) \cdot 2}{\not2} \cdot \not60 \\ S_{60}=(200+59 \cdot 2) \cdot 30 \\ S_{60}=(200+118) \cdot 30 \\ S_{60}=318 \cdot 30 \\ S-{60}= 9540$
$\displaystyle c).a_7=17,~a_2=2,~n=50 \\ a_7=17 \Rightarrow a_{7-1}+r=17 \Rightarrow a_6+r=17 \Rightarrow a_1+6r=17 \Rightarrow \\ \Rightarrow a_1=17-6r \\ a_2=2 \Rightarrow a_{2-1}+r =2 \Rightarrow a_1+r=2 \Rightarrow 17-6r+r= 2 \Rightarrow \\ \Rightarrow -6r+r=2-17 \Rightarrow -5r=-15 \Rihtarrow r= \frac{15}{5} \Rightarrow r=3 \\ a_1=17-6r \Rightarrow a_1=17-6 \cdot 3 \Rightarrow a_1=17-18 \Rightarrow a_1=-1$
$\displaystyle S_{50}= \frac{2 \cdot (-1)+(50-1) \cdot 3}{\not2} \cdot \not50 \\ S_{50}=(-2+49 \cdot 3) \cdot 25 \\ S_{50}=(-2+147) \cdot 25 \\ S_{50}=145 \cdot 25 \\ S_{50}=3625$
$\displaystyle d).a_3=-12,~a_5=36,~n=20 \\ a_3=-12 \Rightarrow a_{3-1}+r=-12 \Rightarrow a_2+r=-12 \Rightarrow a_1+2r=-12 \Rightarrow \\ \Rightarrow a_1=-12-2r \\ a_5=36 \Rightarrow a_{5-1}+r=36 \Rightarrow a_4+r=36 \Rightarrow a_1+4r=36 \Rightarrow \\ \Rightarrow -12-2r+4r=36 \Rightarrow -12+2r=36 \Rightarrow 2r=36+12 \Rightarrow 2r=48 \Rightarrow \\ \Rightarrow r=24 \\ a_1=-12-2r \Rightarrow a_1=-12-2 \cdot 24 \Rightarrow a_1=-12-48 \Rightarrow a_1=-60$
$\displaystyle S_{20}= \frac{2 \cdot (-60)+(20-1) \cdot 24}{\not 2} \vdot \not 20 \\ S_{20}=(-120+19 \cdot 24) \cdot 10 \\ S_{20}=(-120+456) \cdot 10 \\ S_{20}=336 \cdot 10 \\ S_{20}=3360$