Plug in n = 1 into the nth term formula
a(n) = 4n-1
a(1) = 4*1-1
a(1) = 3
So the first term is 3
The second term will be 7 because we add on 4 each time, as indicated by the slope of 4. This is also known as the common difference.
So the nth term is found by adding 4 to the (n-1)st term, in other words,
a(n) = a(n-1)+4
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In summary, the answer is
a1=3; an=an-1+4
which is choice B
Answer:
Can you provide a photo of the reflection please
Step-by-step explanation:
Solution-
First 9 letters of alphabet are- A,B,C,D,E,F,G,H,I
Total number of ways of selection of 4 letters from 9 alphabets = 9C4
=9!÷((4!).(9-4)!) = 9!÷(4!×5!) = (9×8×7×6)÷(24) = 126
The number of ways of arranging these 4 numbers = 4! = 24
∴ Total number of possible permutations = 9C4×4! = 126×24 = 3024
∴ option number 2 is correct.
The answer is a) 1500 because when you divide 675 (the given number of students that cheated on 1 to 7 courses) by 1500 you get .45 or 45%.
