# 1)Find the sixth term of the geometric sequence for which a1=5 and r=3

a.1215
b.3645
c.9375
d.23

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

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

4)Find the sum of the geometric series 128-64+32-_____to 8 terms
a.85
b.255
c.86
d.85/2

5)Find 6 on top of sigma notation n=1 on the bottom 5(-4)n-1 on the right side.
a.6825
b.-4095
c.-1023
d.-5120

## 1) To find the sixth term of a geometric sequence, we use the formula an = a1 * r^(n-1), where an is the nth term, a1 is the first term, r is the common ratio, and n is the term number.

In this case, a1 = 5, r = 3, and we need to find a6.
Plugging these values into the formula:
a6 = 5 * 3^(6-1)
a6 = 5 * 3^5
a6 = 5 * 243
a6 = 1215

2) To find the equation for the nth term of a geometric sequence, we use the formula an = a1 * r^(n-1), where an is the nth term, a1 is the first term, r is the common ratio, and n is the term number.
In the given sequence, the first term a1 = -12 and the common ratio r = 4/(-12) = -1/3.
The equation for the nth term is therefore: an = -12 * (-1/3)^(n-1)

3) To find four geometric means between two numbers, we use the formula for the nth term of a geometric sequence.
Given the first term a1 = 5, and the last term a6 = 1215 (found in question 1), we need to find the geometric mean for each pair of consecutive terms starting from a1.
The equation for the nth term is an = a1 * r^(n-1), where an is the nth term, a1 is the first term, r is the common ratio, and n is the term number.
The common ratio r can be found by taking the nth root of the ratio of two consecutive terms.
In this case, a1 = 5, a6 = 1215, and we want to find the 2nd, 3rd, 4th, and 5th terms.
We can set up the equation a2 = 5 * r^(2-1), a3 = 5 * r^(3-1), a4 = 5 * r^(4-1), and a5 = 5 * r^(5-1), and solve for r using the fact that the ratio of consecutive terms is constant.
By solving these equations simultaneously, we find r = -3.
Substituting this value of r back into the equations, we get the geometric means as -15, 45, -135, and 405.
Adding 6 to each of these values, we get the four geometric means between 5 and 1215 as -9, 51, -129, and 411.

4) To find the sum of a finite 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 the given series, a1 = 128, r = -1/2, and we need to find the sum of the first 8 terms.
Plugging these values into the formula:
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 * 255/384
S = 85

5) The expression "6 on top of sigma notation n=1 on the bottom" represents the sum of a series.
The given series is 5 * (-4)^(n-1).
To find the sum of this series from n = 1 to n = 6, we substitute these values into the series and add them up.
Plugging in n = 1, we get the first term as 5 * (-4)^(1-1) = 5 * (-4)^0 = 5 * 1 = 5.
Plugging in n = 2, we get the second term as 5 * (-4)^(2-1) = 5 * (-4)^1 = 5 * (-4) = -20.
Similarly, we can find the third, fourth, fifth, and sixth terms as -80, 320, -1280, and 5120 respectively.
Adding up these terms: 5 + (-20) + (-80) + 320 + (-1280) + 5120 = -4095.
Therefore, the answer is b) -4095.

## 1) The formula to find the nth term of a geometric sequence is: an = a1 * r^(n-1)

Given that a1 = 5 and r = 3, we can substitute these values into the formula to find the sixth term:
a6 = 5 * 3^(6-1)
a6 = 5 * 3^5
a6 = 5 * 243
a6 = 1215

Therefore, the answer is a. 1215.

2) To find the equation for the nth term of a geometric sequence, we can use the formula: an = a1 * r^(n-1)
Given that the first term, a1, is -12, and the common ratio, r, is -1/3, we can substitute these values into the formula:
an = -12 * (-1/3)^(n-1)

Therefore, the answer is d. an = -12(-1/3)^(n-1).

3) To find the geometric means between two numbers, we can use the formula: G = sqrt(a1 * a2)
Given that the two numbers are 5 and 1215, we can find the geometric means as follows:
G1 = sqrt(5 * 1215)
G2 = sqrt(5 * G1)
G3 = sqrt(5 * G2)
G4 = sqrt(5 * G3)

Therefore, the answer is d. ±247, 489, ±731, 973.

4) The formula to find the sum of a finite geometric series is: Sn = a * (1 - r^n) / (1 - r)
Given that the first term, a, is 128, the common ratio, r, is -1/2, and the number of terms, n, is 8, we can substitute these values into the formula:
Sn = 128 * (1 - (-1/2)^8) / (1 - (-1/2))

Simplifying the expression:
Sn = 128 * (1 - 1/256) / (1 + 1/2)
Sn = 128 * (255/256) / (3/2)
Sn = 128 * 255 * 2 / (256 * 3)
Sn = 1280 / 3
Sn = 85

Therefore, the answer is a. 85.

5) The given expression can be written in sigma notation as follows:
∑(n=1 to 6) 5 * (-4)^(n-1)

We can evaluate this expression by plugging in the values for n from 1 to 6 and summing them up:
n=1: 5 * (-4)^(1-1) = 5 * (-4)^0 = 5 * 1 = 5
n=2: 5 * (-4)^(2-1) = 5 * (-4)^1 = 5 * (-4) = -20
n=3: 5 * (-4)^(3-1) = 5 * (-4)^2 = 5 * 16 = 80
n=4: 5 * (-4)^(4-1) = 5 * (-4)^3 = 5 * (-64) = -320
n=5: 5 * (-4)^(5-1) = 5 * (-4)^4 = 5 * 256 = 1280
n=6: 5 * (-4)^(6-1) = 5 * (-4)^5 = 5 * (-1024) = -5120