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
Answer:
(x² − 2x + 2) (x − 1)
Step-by-step explanation:
Use grouping.
x³ − 3x² + 4x − 2
x³ − 3x² + 2x + 2x − 2
x (x² − 3x + 2) + 2 (x − 1)
x (x − 2) (x − 1) + 2 (x − 1)
(x (x − 2) + 2) (x − 1)
(x² − 2x + 2) (x − 1)
Answer:
32, 46
Step-by-step explanation:
Remember, a is congruent to b modulo d if d divides a-b.
Now, the problem says that b=4 and d=14.
Let a=32. Observe that a-b=28 and 
, then 32 is congruent to 4 modulo 14.
Let a=46. Observe that a-b=46-4=42 and 
, then 46 is congruent to 4 modulo 14.
The sum of each group of numbers are as follows:
1. 110 + 83 + 328 = 521
2. 92 + 37 + 14 + 66 = 209
3. 432 + 11 + 157 + 30 = 630
It does not matter how much you re-arrange the numbers in each group, you will keep arriving at the same answers, this is because the sum of a particular set of numbers will always remain the same.
