1)Find a1 in a geometric series for which Sn=300,r=-3,and n=4
A)15
B)15/2
C)-15
D)1/15
I chose A
5)Find the fifth term of the sequence in which a1=-3,and aN+1=3aN-n
A)-301
B)-99
C)-193
D)-341
I don't know
Jon, for #1 the equation is
300 = a((-3)^4 - 1)/(-3-1)
300 = a(80)/-4
a = -15
since there are only 4 terms, this is easy to check
-15 + 45 + (-135) + 405 = 300
for you #2, I will assume you meant
a1 = -3
aN+1 = 3N - n
then:
a1 = -3
a2 = 3a1 - 2 = -11
a3 = 3a2 - 3 = -36
a4 = 3a3 - 4 = -112
a5 = 3a4 - 5 = -341 which is choice D
Oh, looks like algebra got a little too serious for you. Don't worry, I'm here to clown around and help you out!
For the first question, we are given the sum of the geometric series (Sn), the common ratio (r), and the number of terms (n). We need to find the first term (a1). Let's use a little math and humor to solve this!
The formula for the sum of a geometric series is Sn = a1 * (1 - r^n) / (1 - r). By substituting the given values, we can solve for a1.
Plugging in Sn = 300, r = -3, and n = 4 into the formula, we get:
300 = a1 * (1 - (-3)^4) / (1 - (-3))
Now let's simplify this:
300 = a1 * (1 - 81) / (1 + 3)
300 = a1 * (-80/4)
300 = -20a1
a1 = 300 / -20
a1 = -15
And there you have it! The first term (a1) in this geometric series is -15. So my clown answer for this question would be C) -15!
Now, onto the second question. We need to find the fifth term of the sequence using the recursive formula aN+1 = 3aN - n. Let the clowning commence!
The given value for a1 is -3. To find the fifth term (a5), we can use the recursive formula to calculate each term up to a5.
a2 = 3a1 - 1 = 3(-3) - 1 = -10
a3 = 3a2 - 2 = 3(-10) - 2 = -32
a4 = 3a3 - 3 = 3(-32) - 3 = -99
a5 = 3a4 - 4 = 3(-99) - 4 = -301
So, using my clown calculations and some funny math, the fifth term (a5) of this sequence is -301. Therefore, the clown answer for this question would be A) -301!
Just remember, math can be a tricky circus, but with a bit of humor, we can make it fun and solve any problem. Keep clownin' around!
To find the value of a₁ in a geometric series, we can use the formula:
Sₙ = a₁ * (1 - rⁿ) / (1 - r)
where Sₙ is the sum of the first n terms, a₁ is the first term, r is the common ratio, and n is the number of terms.
Question 1:
Given Sn = 300, r = -3, and n = 4.
Substituting these values into the formula, we get:
300 = a₁ * (1 - (-3)⁴) / (1 - (-3))
Simplifying the equation, we have:
300 = a₁ * (1 - 81) / 4
Multiplying both sides by 4, we get:
1200 = a₁ * (-80)
Dividing both sides by -80, we find:
a₁ = -15
Therefore, the correct answer is C) -15.
Question 5:
To find the fifth term of the sequence, we need to use the recursive formula:
aₙ₊₁ = 3aₙ - n
Given a₁ = -3, we can generate the sequence until we find the fifth term.
a₂ = 3a₁ - 1 = 3(-3) - 1 = -10
a₃ = 3a₂ - 2 = 3(-10) - 2 = -32
a₄ = 3a₃ - 3 = 3(-32) - 3 = -99
a₅ = 3a₄ - 4 = 3(-99) - 4 = -301
Therefore, the correct answer is A) -301.
find a1 for given geometric series.
Sn= 405, r=3, n=4