Answer:
Recursive rule for arithmetic sequence = an = a[n-1] + 3
Step-by-step explanation:
Given arithmetic sequence;
-7, -4, -1, 2, 5, …
Find:
Recursive rule for arithmetic sequence;
Computation:
Let a1 = -7
So,
⇒ a2 = a1 + 3 = -4
⇒ a3 = a2 + 3 = -1
⇒ a4 = a3 + 3 = 2
⇒ a5 = a4 + 3 = 5
So, the recursive formula is
⇒ an = a[n-1] + 3
Recursive rule for arithmetic sequence = an = a[n-1] + 3
The answer i came up with is 0.68 i don't know if thats what you need or not. hopefully i helped.
You have to look at the equation and find a common factor bettween all parts. SO, we see they all have an x, so that is the first thing you can factor out. They also have
2x(5x2 4x -3) C
This is a simple problem based on combinatorics which can be easily tackled by using inclusion-exclusion principle.
We are asked to find number of positive integers less than 1,000,000 that are not divisible by 6 or 4.
let n be the number of positive integers.
∴ 1≤n≤999,999
Let c₁ be the set of numbers divisible by 6 and c₂ be the set of numbers divisible by 4.
Let N(c₁) be the number of elements in set c₁ and N(c₂) be the number of elements in set c₂.
∴N(c₁) =

N(c₂) =

∴N(c₁c₂) =

∴ Number of positive integers that are not divisible by 4 or 6,
N(c₁`c₂`) = 999,999 - (166666+250000) + 41667 = 625000
Therefore, 625000 integers are not divisible by 6 or 4