Question

progresia aritmetica (an) n>1 de ratie r este definita prin anumite elemente date.

4a2 - a4 = 26, s5 = 35. calculati a1 si r

Answer 1

$\displaystyle\bf\\\boxed{\displaystyle\bf NOTIUNI~TEORETICE}\\\\\boxed{\displaystyle\bf 1.}~a_n = a_{1}+(n-1)r,~\forall n\in \mathbb N,~ n\geq 2.\\\\\boxed{\displaystyle\bf 2.}~r=a_n-a_{n-1},~\forall n\in \mathbb N,~n\geq 2.\\\\\boxed{\displaystyle\bf 3.}~a_n = a_{n-1}+r,~\forall n\in \mathbb N,n~\geq 2.\\\\\boxed{\displaystyle\bf 4.}~S_n=\frac{(a_1+a_n)n}{2} ,~\forall n\in \mathbb N,~n\geq 2.\\\\\boxed{\displaystyle\bf 5.}~a_n=\frac{a_{n-1}+a_{n+1}}{2}.\\---------------------------------$

$\displaystyle\bf\\acum~sa~trecem~la~problema.\\a_2=a_1+r,~a4=a1+3r,~S_5=\frac{(a_1+a_5)5}{2},~dar~a_5=a_1+4r,~deci~\\S_5=\frac{(2a_1+4r)5}{2} =5a_1+10r.\\4a_2-a_4=4(a_1+r)-a_1-3r=4a_1+4r-a_1-3r=3a_1+r.\\acum~putem~forma~un~sistem~de~doua~ecuatii~cu~necunoscutele~a_1~si~r.\\\left \{ {{3a_1+r=26|\cdot (-10)} \atop {5a_1+10r=35}} \right. \Leftrightarrow \left \{ {{-30a_1-10r=-260} \atop {5a_1+10r=35}} \right.\\adunam~ecuatiile~\implies -25a_1=-225 \implies a_1=9.\\de~unde~r=-1.$