Determine if the sequence is arithmetic. If it is, write its formaula in both explicit and recursive
notation.
1) 36, 6, -24, -54,... 2) -26, 174, 374, 574,.... 3) 11, 111, 211, 311, .... 4) 11, -9, -29,
-49,....
5) -6, -206, -406, -606 6) 28,-172, -372, -572,....
Write the formula for the arithmetic sequence in both explicit and recursive notation.
7) a19 = -158 and a30 = -246 8) a12 = 71 and a 34 = 291
Solution
1) sequence is arithmetic because common difference is constant d=-30
implicit notation a[n]=36+(-30)*(n-1)
recursive notation a[n]=a[n-1]+(-30)
2)sequence is arithmetic because common difference is constant d=200
implicit notation a[n]=-26+(200)*(n-1)
recursive notation a[n]=a[n-1]+(200)
3)sequence is arithmetic because common difference is constant d=100
implicit notation a[n]=11+(100)*(n-1)
recursive notation a[n]=a[n-1]+(100)
4)sequence is arithmetic because common difference is constant d=-20
implicit notation a[n]=11+(-20)*(n-1)
recursive notation a[n]=a[n-1]+(-20)
5)sequence is arithmetic because common difference is constant d=-200
implicit notation a[n]=-6+(-200)*(n-1)
recursive notation a[n]=a[n-1]+(-200)
6)sequence is arithmetic because common difference is constant d=-200
implicit notation a[n]=28+(-200)*(n-1)
recursive notation a[n]=a[n-1]+(-200)
7)a19=-158 and a30=-246
a19=a1+18*d
a30=a1+29*d
subtracting a19 from a30 we have
a30-a19=-246-(-158)
11*d=-88
d=-8
putting d=-8 in a19 =a1+18*d we get a1=-14
implicit notation a[n]=-14+(-8)*(n-1)
recursive notation a[n]=a[n-1]+(-8)
8)a12=71 and a34=291
a12=a1+11*d
a34=a1+33*d
subtracting a12 from a34 we have
a34-a12=-291-(71)
22*d=220
d=10
putting d=10 in a12 =a1+11*d we get a1=-39
implicit notation a[n]=-39+(10)*(n-1)
recursive notation a[n]=a[n-1]+(10).
Ride the Storm: Navigating Through Unstable Periods / Katerina Rudko (Belka G...
Determine the antiderivative of 1(x+5)SolutionWell calculat.pdf
1. Determine the antiderivative of 1/(x+5)
Solution
We'll calculate the antiderivative F(x) integrating the given function.
F(x) is a function such as dF/dx = f(x).
Int f(x)dx = F(x) + C
Int dx/(x+5)
We'll substitute the denominator of the function by another variable:
x + 5 = t
We'll differentiate both sides:
(x+5)'dx = dt
dx = dt
We'll re-write the indefinite integral using the variable t:
Int dt/t = ln |t| + C
We'll substitute t by the expression in x:
F(x) = ln |x+5| + C
The antiderivative of the function y = 1/(x+5) is F(x)=ln |x+5| + C.
