Question

Sa se determine suma primilor n termeni ai progresiei aritmetice a indice n daca:

a) a indice 7=17 , a indice 2=2 , n=50
b) a indice 3= -12 , a indice 5=36 , n=20

Answer 1

$\displaystyle a).a_7=17,~a_2=2,~n=50~~~~~~~~~~~~~~~~~~~~~~~\boxed{S_n= \frac{2a_1+(n-1) \cdot r}{2} \cdot n} \\ a_7=17 \Rightarrow a_{7-1}+r=17 \Rightarrow a_6+r=17 \Rightarrow a_1+6r=17 \Rightarrow \\ \Rightarrow a_1=17-6 r \\ a_2=2 \Rightarrow a_{2-1}+r=2 \Rightarrow a_1+r=2 \Rightarrow 17-6r+r=2 \Rightarrow \\ \Rightarrow 17-5r=2 \Rightarrow -5r=2-17 \Rightarrow -5r=-15 \Rightarrow r=3 \\ a_1=17-6r \Rightarrow a_1=17-6 \cdot 3 \Rightarrow a_1=17-18 \Rightarrow a_1=-1$
$\displaystyle S_{50}= \frac{-2+49 \cdot3}{2} \cdot 50 \\ S_{50}= \frac{-2+147}{2} \cdot 50 \\ S_{50}= \frac{145}{2} \cdot 50 \\ S_{50}=145 \cdot 25 \\ S_{50}=3625$
$\displaystyle b).a_3=-12,~a_5=36,~n=20 \\ a_3=-12 \Rightarrow a_{3-1}+r=-12 \Rightarrow a_2+r=-12 \Rightarrow a_1+2r=-12 \Rightarrow \\ \Rightarrow a_1=-12-2r \\ a_5=36 \Rightarrow a_{5-1}+r=36 \Rightarrow a_4+r=36 \Rightarrow a_1+4r=36 \Rightarrow \\ \Rightarrow -12-2r+4r=36 \Rightarrow -12+2r=36 \Rightarrow 2r=36+12 \Rightarrow 2r=48 \Rightarrow \\ \Rightarrow r=24 \\ a_1=-12-2r \Rightarrow a_1=-12-2 \cdot 24 \Rightarrow a_1=-12-48 \Rightarrow a_1=-60$
$\displaystyle S_{20}= \frac{-120+19 \cdot 24}{2} \cdot 20 \\ S_{20}= \frac{-120+456}{2} \cdot 20 \\ S_{20}= \frac{336}{2} \cdot 20 \\ S_{20}=168 \cdot 20 \\ S_{20}=3360$