a(1)=3; a(2)=2
d=a(n+1)-a(n) = a(2)-a(1) = 2-3 = -1.
a(n)=a1+d*(n-1) => a(3)=a(1)+(-1)*(3-1)=3-1*(3-1)=3-3+1=1
a(4)=3-1*(4-1)=0 =>
a(3)*a(4)=1*0=0.
ответ: 0.
a(1)=3; a(2)=2
d=a(n+1)-a(n) = a(2)-a(1) = 2-3 = -1.
a(n)=a1+d*(n-1) => a(3)=a(1)+(-1)*(3-1)=3-1*(3-1)=3-3+1=1
a(4)=3-1*(4-1)=0 =>
a(3)*a(4)=1*0=0.
ответ: 0.