1. The geometric mean between the first two terms in a geometric sequence is 32. If the third term is 4, find the first term.

2. Insert a geometric mean between k and 1/k.
3. If 2 and 3 are two geometric means between m and n, fond the values of m and n.
4. Three positive numbers from a geometric sequence. If the geomeyric mean of the first two numbers is 6 and tbe geometric mean of the last two numbers is 24, find the three numbers and their common ratio.

The geometric means between the first twi terms and geometric sequence is 32 the third term is 4 find the first term

Insert one geometric mean between 81 and 100.

1. To find the first term in a geometric sequence, we can use the formula for the nth term of a geometric sequence:

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 position of the term.

Given that the geometric mean between the first two terms is 32, we can write:

sqrt(A1 * (A1 * r)) = 32

Simplifying this equation, we get:

sqrt(A1^2 * r) = 32

Squaring both sides, we get:

A1^2 * r = 32^2

A1^2 * r = 1024

Since we know that the third term is 4, we can substitute these values into the equation to solve for the first term:

A1^2 * r^2 = 4

Substituting A1^2 * r from the previous equation, we get:

1024 * r = 4

Solving this equation for r, we find:

r ≈ 0.00390625

Now, substitute this value back into A1^2 * r = 1024 to solve for A1:

A1^2 * 0.00390625 = 1024

A1^2 ≈ 262144

Taking the square root of both sides, we find:

A1 ≈ 512

Therefore, the first term of the geometric sequence is approximately 512.

2. To insert a geometric mean between k and 1/k, we need to find the value of the geometric mean.

The geometric mean between two numbers can be found using the formula:

Geometric Mean = sqrt(First Number * Second Number)

In this case, the first number is k and the second number is 1/k. Substituting these values into the formula, we get:

Geometric Mean = sqrt(k * (1/k))

Simplifying this expression, we get:

Geometric Mean = sqrt(1)

Since the square root of 1 is 1, the value of the geometric mean is 1.

Therefore, to insert a geometric mean between k and 1/k, we can simply write 1.

3. To find the values of m and n when 2 and 3 are two geometric means between them, we can use the formula for the nth term of a geometric sequence:

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 position of the term.

Given that 2 is the first geometric mean and 3 is the second geometric mean, we can represent the sequence as:

m, 2, 3, n

We know that 2 is the first geometric mean, so:

2 = m * r

Similarly, we know that 3 is the second geometric mean, so:

3 = 2 * r
=> 3 = (m * r) * r
=> 3 = m * r^2

Now, we can solve the two equations simultaneously to find the values of m and n. Divide the second equation by the first equation:

3/2 = r

Substituting this value of r into the first equation:

2 = m * (3/2)
=> 2 = (3/2) * m
=> 2 = (3m)/2
=> 4 = 3m
=> m = 4/3

Therefore, the value of m is 4/3. To find the value of n, we can substitute m = 4/3 and r = 3/2 into the equation 3 = m * r^2:

3 = (4/3) * (3/2)^2

Simplifying this expression, we get:

3 = (4/3) * (9/4)
=> 3 = 3

Therefore, the value of n is 3.

Hence, m = 4/3 and n = 3.

4. Let the three positive numbers in the geometric sequence be A, Ar, and Ar^2.

From the given information, we know that the geometric mean of the first two numbers (A and Ar) is 6:

sqrt(A * (Ar)) = 6

Simplifying this equation, we get:

sqrt(A^2 * r) = 6

Squaring both sides, we get:

A^2 * r = 36

Similarly, we know that the geometric mean of the last two numbers (Ar and Ar^2) is 24:

sqrt(Ar * (Ar^2)) = 24

Simplifying this equation, we get:

sqrt(Ar^3) = 24

Squaring both sides, we get:

Ar^3 = 576

Now we have a system of two equations with two variables (A and r). We can solve them to find the values of A, r, and the three numbers.

First, let's solve the equation Ar^3 = 576 for Ar:

Ar^3 = 576
Ar = 6

Substituting this value of Ar into the equation A^2 * r = 36, we get:

(6r)^2 * r = 36
36r^3 = 36

Dividing both sides by 36, we get:

r^3 = 1

Taking the cube root of both sides, we find:

r = 1

Now, substituting r = 1 back into the equation Ar = 6, we get:

A = 6

So, the three numbers in the geometric sequence are 6, 6, and 6, and their common ratio is 1.

Find the sum of the terms of geometric sequence where the first term is 4, the last term is 324 and the common ratio is 3.

16 and 81 [3]

looks like you expect somebody to do an assignment or homework for you, and you have not shown any work or indication where your problems lie.

What don' t you understand about these questions.
They all seem to deal with "geometric mean"

btw, you will have more luck posting individual problems, showing some steps or work on your part.