1)Find the sixth term of the geometric sequence for which a1=5 and r=3
A)1215
B)3645
C)9375
D)23
I chose A
2)Write an equation for the nth term of the geometric sequence -12,4,-4/3...
A)aN=-12(1/3)n-1
B)aN=12(-1/3)n-1
C)aN=-12(-1/3)-n+1
D)aN=-12(-1/3)n-1
I chose C
3)Find four geometric means between 5 and 1215.
A)+-15,45,+-135,405
B)15,45,135,405
C)247,489,731,973
D)+-247,489,+-731,973
I chose D
4)Find the sum of the geometric series 128-64+32-______to 8 terms
A)85
B)255
C)86
D)85/2
I chose A
5)Sigma again.
above the sigma sign there is a 6 and below is n=1.To the right is 5(-4)n-1. The question is find.
A)6825
B)-4095
C)-1023
D)-5120
I chose B
All are correct except #2 and #3
#2:
a+-12, r = -1/3)
so it has to be D, I don't understand why you would choose C, the exponent does not follow the formula.
#3:
the sequence would be 5, 5r, 5r^2, 5r^3, 5r^4, 1215
so from the last 3 terms:
1215/5r^4 = 5r^4/5r^3
.
.
.
r^5=243
r=+3
so the terms in between would be 15, 45, 135, and 450 which is B
yeah for #2 i did pick D sorry for the typo
1) To find the sixth term of a geometric sequence, we can use the formula an = a1 * r^(n-1), where a1 is the first term, r is the common ratio, and n is the term number.
Given that a1 = 5 and r = 3, we substitute these values into the formula:
a6 = 5 * 3^(6-1)
a6 = 5 * 3^5
a6 = 5 * 243
a6 = 1215
Therefore, the correct answer is A) 1215.
2) To write an equation for the nth term of a geometric sequence, we can use the formula an = a1 * r^(n-1).
Given the geometric sequence -12, 4, -4/3, we know that the first term a1 = -12. In order to find the common ratio, we can divide any term by the previous term. In this case, (-4/3) / 4 = -1/3. Therefore, the common ratio r = -1/3.
Substituting these values into the formula, we get:
aN = -12 * (-1/3)^(n-1)
Therefore, the correct answer is D) aN = -12(-1/3)^(n-1).
3) To find four geometric means between 5 and 1215, we need to find the common ratio (r) of the geometric sequence. The formula to find the common ratio is r = (term2/term1)^(1/(n-1)), where term2 is the second term, term1 is the first term, and n is the total number of terms (including the first and last terms).
Given that term1 = 5, term2 = 1215, and n = 6 (including the first and last terms), we substitute these values into the formula:
r = (1215/5)^(1/(6-1))
r = 243^(1/5)
r = 3
Now, we can find the four geometric means using the formula an = a1 * r^(n-1).
First geometric mean:
a2 = 5 * 3^(2-1)
a2 = 15
Second geometric mean:
a3 = 15 * 3^(3-1)
a3 = 45
Third geometric mean:
a4 = 45 * 3^(4-1)
a4 = 135
Fourth geometric mean:
a5 = 135 * 3^(5-1)
a5 = 405
Therefore, the correct answer is B) 15, 45, 135, 405.
4) To find the sum of a geometric series with a given number of terms, we can use the formula S = a1 * (1 - r^n) / (1 - r), where S is the sum of the series, a1 is the first term, r is the common ratio, and n is the number of terms.
Given the geometric series 128, -64, 32, ..., we can see that the first term a1 = 128 and the common ratio r = -1/2. We are asked to find the sum of the series to 8 terms, so n = 8.
Substituting these values into the formula, we get:
S = 128 * (1 - (-1/2)^8) / (1 - (-1/2))
S = 128 * (1 - 1/256) / (1 + 1/2)
S = 128 * (255/256) / (3/2)
S = (128 * 255 * 2) / (256 * 3)
S = 85
Therefore, the correct answer is A) 85.
5) In the given expression ∑6n=15(-4)^(n-1), we want to find the value.
Using the formula for the sum of a geometric series, we have:
S = a * (1 - r^n) / (1 - r)
In this case, a = 5, r = -4, and n ranges from 1 to 6.
Substituting these values into the formula, we get:
S = 5 * (1 - (-4)^6) / (1 - (-4))
Now, we evaluate this expression:
S = 5 * (1 - 4096) / (1 + 4)
S = 5 * (-4095) / 5
S = -4095
Therefore, the correct answer is B) -4095.
1) To find the sixth term of a geometric sequence, we use the formula an = a1 * r^(n-1), where a1 is the first term and r is the common ratio. In this case, a1 = 5 and r = 3. Plugging in these values, we have a6 = 5 * 3^(6-1) = 5 * 3^5 = 5 * 243 = 1215. Therefore, the answer is A) 1215.
2) To find the equation for the nth term of a geometric sequence, we use the formula an = a1 * r^(n-1), where a1 is the first term and r is the common ratio. In this case, the first term is -12 and the common ratio is -1/3. Plugging in these values, we have an = -12 * (-1/3)^(n-1). Simplifying this equation, we get aN = -12(-1/3)^(n-1), which corresponds to option D) aN = -12(-1/3)n-1.
3) To find four geometric means between 5 and 1215, we need to find the common ratio (r) of the sequence. We can do this by taking the nth root of the ratio between the 2nd term and the 1st term. In this case, the 2nd term is 1215 and the 1st term is 5. Thus, the common ratio is (1215/5)^(1/6) ≈ 3. Plugging in the values, we get the sequence: 5, 5 * 3, 5 * 3^2, 5 * 3^3, 5 * 3^4, 5 * 3^5. Simplifying and expanding, we have 5, 15, 45, 135, 405, 1215. Therefore, the answer is D) ±247, 489, ±731, 973.
4) To find the sum of a geometric series, we use the formula S = a1 * (1 - r^n) / (1 - r), where S is the sum, a1 is the first term, r is the common ratio, and n is the number of terms. In this case, a1 = 128, r = -1/2, and n = 8. Plugging in these values, we have S = 128 * (1 - (-1/2)^8) / (1 - (-1/2)) = 128 * (1 - 1/256) / (1 + 1/2) = 128 * (256/256 - 1/256) / (2/2 + 1/2) = 128 * (255/256) / (3/2) = 128 * (255/256) * (2/3) = 85. Therefore, the answer is A) 85.
5) The expression given above the sigma sign represents a geometric series. It shows the sum of the terms from n = 1 to n = 6. The formula for the nth term of a geometric sequence is aN = a1 * r^(n-1), where a1 is the first term and r is the common ratio. In this case, a1 = 5 and r = -4. Plugging in these values, we have the sum as S = 5 * (-4)^(1-1) + 5 * (-4)^(2-1) + 5 * (-4)^(3-1) + 5 * (-4)^(4-1) + 5 * (-4)^(5-1) + 5 * (-4)^(6-1) = 5 * 1 + 5 * (-4) + 5 * 16 + 5 * (-64) + 5 * 256 + 5 * (-1024) = 5 - 20 + 80 - 320 + 1280 - 5120 = -4095. Therefore, the answer is B) -4095.