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
now, we can find kth term
now, we can plug values
and we get
u(k) must be negative
so,
now, we can solve for k
so, it's closest integer value is
..............Answer
In order to add these fractions we have to make a common denominator.
1/5 can turn into 2/10 Its the same thing. And its easier to solve with.
now lets multiply.
×
Now we add across
2 + 1 = 3
------------
10 + 10 = 10 ( When adding fractions like these the denominator stays the same)
So your answer is...
Good Luck! :)
