1.
( a(n-1) + a(n+1) )÷2 = ( 3(n-1)-4 + 3(n+1)-4 )÷2 = ( 3n-7 + 3n-1 )÷2 = ( 6n-8 )÷2 = 3n-4 = a(n) => a(n) este progresie aritmetica
r = a(2) - a(1) = 3×2-4 - ( 3×1-4 ) = 2+1 = 3
2.
S(n)=n^2 => ( n×( a(1)+a(n) ) )÷2=n^2 => n×( 2×a(1)+(n-1)×r )=2×n^2 => 2×a(1)+(n-1)×r=2×n => r=( 2×n-2×a(1) ) ÷ (n-1) => r nu este nr constant => sirul nu este progresie aritmetica
S(n)=n^2+4 => ( n×( a(1)+a(n) ) )÷2=n^2+4 => n×( 2×a(1)+(n-1)×r )=2×n^2 + 8 => 2×a(1)+(n-1)×r= ( 2×n^2+8 ) ÷n => (n-1)×r=( 2×n^2+8 )÷n - 2×a(1) => r = ( ( 2×n^2+8 )÷n - 2×a(1) ) ÷ (n-1) => r nu este nr constant => sirul nu este progresie aritmetica
