Question

(an) n ≥1= progresie aritmetica , a₁+a₂+a₃ = 6 , a₁²+a₂²+a₃²= 14..a₁=? , r ₁= ? Va rog sa ma ajutati !!

Answer 1

[tex]a_1+a_1+r+a_1+2r=6=>3a_1+3r=6|:3=>a_1+r=2\\ a_1^2+(a_1+r)^2+(a_1+2r)^2=14\\ (2-r)^2+2^2+(2+r)^2=14\\ 4-4r+r^2+4+4r+r^2=10\\ 2r^2=2=>r^2=1=>r=\pm1\\ r=1=>a_1=1\\ r=-1=>a_1=3[/tex]