Now if a 1 is the first term of an arithmetic sequence, an is the nth term of the sequence, and
for each term other than the first term, d is the value of that term minus the term right
before it, then an = a 1 + d(n − 1).You know that a 1 = 8 and a 25 = 104. So you have n = 25. Using the formula an = a 1 + d(n − 1)
with a 1 = 8, a 25 = 104, and n = 25, you have 104 = 8 + d(25 − 1). Then 104 = 8 + 24d, 96 = 24d,
and . Now d is a 2 − a 1 . Thus, a 2 − a 1 = 4.44 . B
For all θ, sin (θ + π) = −sin θ.
For all θ, sin (θ + 2π) = sin θ.So sin θ + sin (θ + π) + sin (2π + θ) = sin θ − sin θ + sin θ = sin θ.45 . E
The definition of x! for positive integers x is x! = x(x − 1)(x − 2)...(3)(2)(1).
(n + 4)! = (n + 4)(n + 3)(n + 2)(n + 1)(n)...(3)(2)(1)(n + 7)! = (n + 7)(n + 6)(n + 5)(n + 4)(n + 3)(n + 2)(n + 1)(n)...(3)(2)(1)To solve this question, write(n + 7)! = (n + 7)(n + 6)(n + 5)(n + 4)!
Then46 . E
When a function n(x) is translated to the right v units, where v > 0, the resulting function, say
r(x), can be described by r(x) = n(x − v).