1) To find the sixth term of a geometric sequence, you can use the formula:

an = a1 * r^(n-1)

Given that a1 = 5 and r = 3, we can substitute these values into the formula:

a6 = 5 * 3^(6-1)

= 5 * 3^5

= 5 * 243

= 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 that the sequence is -12, 4, -4/3, ..., we can see that a1 = -12 and r = 4/(-12) = -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 can use the formula:

gm = sqrt(a * b)

Given that a = 5 and b = 1215, we can find the first geometric mean:

gm1 = sqrt(5 * 1215)

= sqrt(6075)

= 77.98 (approximately)

Then, we can find the second geometric mean by using gm1 as a and b:

gm2 = sqrt(77.98 * 1215)

= sqrt(94622.7)

= 307.93 (approximately)

Similarly, we can find the third and fourth geometric means:

gm3 = sqrt(307.93 * 1215)

= sqrt(374040.95)

= 611.99 (approximately)

gm4 = sqrt(611.99 * 1215)

= sqrt(743536.85)

= 862.87 (approximately)

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

4) To find the sum of a geometric series, we can use the formula:

Sn = a1 * (1 - r^n) / (1 - r)

Given that a1 = 128, r = -1/2, and n = 8, we can substitute these values into the formula:

S8 = 128 * (1 - (-1/2)^8) / (1 - (-1/2))

= 128 * (1 - 1/256) / (3/2)

= 128 * (255/256) / (3/2)

= 128 * (255/256) * (2/3)

= 85

Therefore, the correct answer is (a) 85.

5) The given expression can be written in sigma notation as follows:

Σ(5 * (-4)^(n-1)), from n = 1 to 6

Expanding the series, we get:

5 * (-4)^(1-1) + 5 * (-4)^(2-1) + 5 * (-4)^(3-1) + 5 * (-4)^(4-1) + 5 * (-4)^(5-1) + 5 * (-4)^(6-1)

Simplifying each term, we get:

5 * (-4)^0 + 5 * (-4)^1 + 5 * (-4)^2 + 5 * (-4)^3 + 5 * (-4)^4 + 5 * (-4)^5

= 5 * 1 + 5 * (-4) + 5 * 16 + 5 * (-64) + 5 * 256 + 5 * (-1024)

= 5 - 20 + 80 - 320 + 1280 - 5120

= -4095

Therefore, the correct answer is (b) -4095.

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

a.1215

b.3645

c.9375

d.23

answer=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

answer=d

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

answer=d

4)Find the sum of the geometric series 128-64+32-_____to 8 terms

a.85

b.255

c.86

d.85/2

answer=a

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

answer=b

Step-by-Step Bot
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Explain Bot
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To find the sixth term of a geometric sequence with a1 = 5 and r = 3, we can use the formula for the nth term of a geometric sequence: an = a1 * r^(n-1).

Substituting the given values, we have a6 = 5 * 3^(6-1) = 5 * 3^5 = 5 * 243 = 1215. Thus, the answer is a.

To write an equation for the nth term of the geometric sequence -12, 4, -4/3, ..., we need to determine the common ratio (r). We can do this by dividing any term by its preceding term. In this case, r = (4) / (-12) = -1/3.

Using this value of r, the equation for the nth term is: an = (-12) * (-1/3)^(n-1). Simplifying this expression, we get an = -12 * (-1/3)^(n-1). Thus, the answer is d.

To find four geometric means between 5 and 1215, we need to determine the common ratio (r). In this case, r = √(1215/5) = √243 = ±3.

Using this value of r, the geometric means can be found by multiplying the previous term by the common ratio. Starting with 5, the sequence is 5, ±15, 45, ±135, 405, ±1215. Thus, the answer is d.

To find the sum of the geometric series 128 - 64 + 32 - ..., we can use the formula for the sum of a finite geometric series: S = a * (1 - r^n) / (1 - r), where S is the sum, a is the first term, r is the common ratio, and n is the number of terms.

In this case, a = 128, r = -1/2, and n = 8. Substituting these values, we have S = 128 * (1 - (-1/2)^8) / (1 - (-1/2)) = 128 * (1 - 1/256) / (3/2) = 128 * (255/256) / (3/2) = 128 * 255 / (256 * 3) = 85. Thus, the answer is a.

The expression 6 on top of sigma notation with n = 1 on the bottom and 5(-4)^(n-1) on the right side represents the sum of a geometric series.

Expanding the terms, we have 5(-4)^(1-1) + 5(-4)^(2-1) + 5(-4)^(3-1) + 5(-4)^(4-1) + 5(-4)^(5-1) + 5(-4)^(6-1). Simplifying this expression, we get 5(-4)^0 + 5(-4)^1 + 5(-4)^2 + 5(-4)^3 + 5(-4)^4 + 5(-4)^5 = 5 + (-20) + 80 + (-320) + 1280 + (-5120) = -4095. Thus, the answer is b.

Substituting the given values, we have a6 = 5 * 3^(6-1) = 5 * 3^5 = 5 * 243 = 1215. Thus, the answer is a.

To write an equation for the nth term of the geometric sequence -12, 4, -4/3, ..., we need to determine the common ratio (r). We can do this by dividing any term by its preceding term. In this case, r = (4) / (-12) = -1/3.

Using this value of r, the equation for the nth term is: an = (-12) * (-1/3)^(n-1). Simplifying this expression, we get an = -12 * (-1/3)^(n-1). Thus, the answer is d.

To find four geometric means between 5 and 1215, we need to determine the common ratio (r). In this case, r = √(1215/5) = √243 = ±3.

Using this value of r, the geometric means can be found by multiplying the previous term by the common ratio. Starting with 5, the sequence is 5, ±15, 45, ±135, 405, ±1215. Thus, the answer is d.

To find the sum of the geometric series 128 - 64 + 32 - ..., we can use the formula for the sum of a finite geometric series: S = a * (1 - r^n) / (1 - r), where S is the sum, a is the first term, r is the common ratio, and n is the number of terms.

In this case, a = 128, r = -1/2, and n = 8. Substituting these values, we have S = 128 * (1 - (-1/2)^8) / (1 - (-1/2)) = 128 * (1 - 1/256) / (3/2) = 128 * (255/256) / (3/2) = 128 * 255 / (256 * 3) = 85. Thus, the answer is a.

The expression 6 on top of sigma notation with n = 1 on the bottom and 5(-4)^(n-1) on the right side represents the sum of a geometric series.

Expanding the terms, we have 5(-4)^(1-1) + 5(-4)^(2-1) + 5(-4)^(3-1) + 5(-4)^(4-1) + 5(-4)^(5-1) + 5(-4)^(6-1). Simplifying this expression, we get 5(-4)^0 + 5(-4)^1 + 5(-4)^2 + 5(-4)^3 + 5(-4)^4 + 5(-4)^5 = 5 + (-20) + 80 + (-320) + 1280 + (-5120) = -4095. Thus, the answer is b.