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almond37 [142]
3 years ago
10

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

Mathematics
1 answer:
omeli [17]3 years ago
3 0

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


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