Question

Fie progresia aritmetica (an) cu a2=4 si a5 =19 a)calculati a3 a1 S9

Answer 1

$\displaystyle a_2=4,~a_5=19,~a_3=?,~a_1=?,~S_9=? \\ a_2=4 \Rightarrow a_{2-1}+r=4 \Rightarrow a_1+r=4 \Rightarrow a_1=4-r \\ a_5=19 \Rightarrow a_{5-1}+r=19 \Rightarrow a_4+r=19 \Rightarrow a_1+4r=19 \Rightarrow \\ \Rightarrow 4-r+4r=19 \Rightarrow 4+3r=19 \Rightarrow 3r=19-4 \Rightarrow 3r=15 \Rightarrow \\ \Rightarrow r=\frac{15}{3} \Rightarrow r=5\\a_1=4-r \Rightarrow a_1=4-5 \Rightarrow \boxed{a_1=-1} \\a_3=a_{3-1}+r \Rightarrow a_3=a_2+r\Rightarrow a_3=a_1+2r\Rightarrow a_3=-1+2\cdot5 \Rightarrow$
$\displaystyle \Rightarrow a_3=-1+10 \Rightarrow \boxed{a_3=9} \\ \\ \boxed{S_n= \frac{2a_1+(n-1) \cdot r}{2} \cdot n} \\ \\ S_9= \frac{2 \cdot (-1)+(9-1) \cdot 5}{2} \cdot 9 \\ S_9= \frac{-2+8 \cdot 5}{2}\cdot 9 \\ S_9= \frac{-2+40}{2} \cdot 9 \\ S_9= \frac{38}{2} \cdot 9 \\ S_9=19 \cdot 9 \\ \boxed{S_9=171}$