We are given
arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103
so, first term is 124
u(1)= 124
now, we can find common difference
[tex] d=u(2)-u(1) [/tex]
[tex] d=117-124 [/tex]
[tex] d=-7 [/tex]
now, we can find kth term
[tex] u(k)=u(1)+(k-1)d [/tex]
now, we can plug values
and we get
[tex] u(k)=124+(k-1)*-7 [/tex]
[tex] u(k)=124-7k+7 [/tex]
[tex] u(k)=131-7k [/tex]
u(k) must be negative
so,
[tex] u(k)=131-7k<0 [/tex]
[tex] 131-7k<0 [/tex]
now, we can solve for k
[tex] 7k>131 [/tex]
[tex] k>18.714 [/tex]
so, it's closest integer value is
[tex] k=19 [/tex]..............Answer
