Question

Să se determine suma primilor n termeni ai 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.


Să se determine termenul a10 al progresiei aritmetice (An) daca :
a1=-2, a8 = 12

Answer 1

$\displaystyle 1a).a_1=-1, r=3,~n=12 \\ \\ S_{12}= \frac{2 \cdot (-1)+(12-1) \cdot 3}{2} \cdot 12 \\ \\ S_{12}= \frac{-2+11 \cdot 3}{\not2} \cdot \not12 \\ S_{12}=(-2+33) \cdot 6 \\ S_{12}=31 \cdot 6 \\ S_{12}=186 \\ b).a_1=100,~r=2,~n=60 \\ S_{60}= \frac{2 \cdot 100+(60-1) \cdot 2}{2} \cdot 60 \\ \\ S_{60}= \frac{200+59 \cdot 2}{\not2} \cdot \not60 \\ S_{60}=(200+118)\cdot 30 \\ S_{60}=318 \cdot 30 \\ S_{60}=9540$
$\displaystyle c).a_7=17,a_2=2, ~n=50 \\ a_7=17 \Rightarrow a_{7-1}+r=17 \Rightarrow a_6+r=17 \Rightarrow a_1+6r=17 \Rightarrow \\ \Rightarrow a_1=17-6 r \\ a_2=2 \Rightarrow a_{2-1}+r=2 \Rightarrow a_1+r=2 \Rightarrow 17-6r+r=2 \Rightarrow \\ \Rightarrow 17-5r=2 \Rightarrow -5r=2-17 \Rightarrow -5r=-15 \Rightarrow r=3 \\ a_1=17-6r \Rightarrow a_1=17-6 \cdot 3 \Rightarrow a_1=17-18 \Rightarrow a_1=-1$
$\displaystyle S_{50}= \frac{2 \cdot (-1)+(50-1) \cdot 3}{2} \cdot 50 \\ \\ S_{50}= \frac{-2+49 \cdot3}{\not2} \cdot \not50 \\ \\ S_{50}=(-2+147) \cdot 25 \\ S_{50}=145 \cdot 25 \\ S_{50}=3625$
[tex]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 [/tex]
$\displaystyle S_{20}= \frac{2 \cdot (-60)+(20-1) \cdot 24}{2} \cdot 20 \\ \\ S_{20}= \frac{-120+19 \cdot 24}{\not2} \cdot \not20 \\ \\ S_{20}=(-120+456) \cdot 10 \\ S_{20}=336 \cdot 10 \\ S_{20}=3360$$\displaystyle 2).a_1=-2,~a_8=12,~a_{10}=? \\ a_8=12 \Rightarrow a_{8-1}+r=12\Rightarrow a_7+r=12 \Rightarrow a_1+7r=12 \Rightarrow \\ \Rightarrow -2+7r=12 \Rightarrow 7r=12+2 \Rightarrow 7r=14 \Rightarrow r= \frac{14}{7} \Rightarrow r=2 \\ a_{10}=a_{10-1}+r \Rightarrow a_{10}=a_9+r \Rightarrow a_{10}=a_1+9r \Rightarrow \\ \Rightarrow a_{10}=-2+9 \cdot 2 \Rightarrow a_{10}=-2+18 \Rightarrow a_{10}=16$