Question

Given the arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103,...

Answer 1

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

$d=u(2)-u(1)$

$d=117-124$

$d=-7$

now, we can find kth term

$u(k)=u(1)+(k-1)d$

now, we can plug values

and we get

$u(k)=124+(k-1)*-7$

$u(k)=124-7k+7$

$u(k)=131-7k$

u(k) must be negative

so,

$u(k)=131-7k$

$131-7k$

now, we can solve for k

$7k>131$

$k>18.714$

so, it's closest integer value is

$k=19$..............Answer