1. Find the common ratio and the 7th term of the sequence 2/9, 2/3, 2, 6, 18, .....
2. The 3rd and 5th terms of a geometric sequence are 2 and 1/2 respectively. What is the first term?
3. Find the geometric mean between:
a. 1/2 and 1/3
b. sqrt5 - sqrt2
common ratio = r = 6/2 = 3 (I took the easiest fraction)
first term = a = 2/9
term7 = ar^6 = (2/9)(3^6) = 162
2. ar^2 = 2
ar^4 = 1/2
divide the 2nd equation by the first
r^2 = 1/4
r = ± 1/2
then in ar^2 = 2
a(1/4) = 2
a = 8
3. a)
x/(1/2) = (1/3)/x
x^2 = 1/6
x = ± 1/√6
b) check your typing
is it √5 and -√2 ?
9
1. To find the common ratio of a geometric sequence, we can divide any term by its previous term.
Let's divide each term by the previous term:
(2/3) / (2/9) = (2/3) * (9/2) = 3
Similarly, dividing each term by its previous term, we find that:
2 / (2/3) = 3
6 / 2 = 3
18 / 6 = 3
So, the common ratio of the sequence is 3.
To find the 7th term, 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 term number.
Given that a1 = 2/9 and r = 3, we can find the 7th term:
a7 = (2/9) * 3^(7-1) = (2/9) * 3^6 = (2/9) * 729 = 162/9 = 18
Therefore, the 7th term of the sequence is 18.
2. In a geometric sequence, the common ratio can be found by dividing any term by its previous term.
Let's divide the 3rd term by the 1st term:
2 / a = 2
Simplifying this equation, we find that a = 1.
Therefore, the first term of the geometric sequence is 1.
3.
a. To find the geometric mean between 1/2 and 1/3, we can take the square root of their product:
√((1/2) * (1/3)) = √(1/6) = 1/√6
Therefore, the geometric mean between 1/2 and 1/3 is 1/√6.
b. To find the geometric mean between √5 and √2, we can take the square root of their product:
√(√5 * √2) = √(√10)
Therefore, the geometric mean between √5 and √2 is √(√10).
1. To find the common ratio (r) of a geometric sequence, you divide two consecutive terms. In this case, let's divide the second term (2/3) by the first term (2/9):
(2/3) ÷ (2/9) = (2/3) × (9/2) = 3/1 = 3
Therefore, the common ratio (r) is 3.
To find the 7th term of the sequence, you can use the formula for the nth term of a geometric sequence:
nth term = a * r^(n-1)
where a is the first term and r is the common ratio.
In this case, the first term (a) is 2/9, the common ratio (r) is 3, and we want to find the 7th term (n = 7). Plugging these values into the formula:
7th term = (2/9) * 3^(7-1)
= (2/9) * 3^6
= (2/9) * 729
= 162/3
= 54
Therefore, the 7th term of the sequence is 54.
2. In this problem, you are given the 3rd term (2) and the 5th term (1/2) of a geometric sequence.
Let's assume the first term is a and the common ratio is r.
According to the problem, the 3rd term is 2:
a * r^(3-1) = 2
a * r^2 = 2
Similarly, for the 5th term, we have:
a * r^(5-1) = 1/2
a * r^4 = 1/2
Now, we have a system of equations:
a * r^2 = 2 ---- (1)
a * r^4 = 1/2 ---- (2)
Dividing equation (2) by equation (1):
(r^4) / (r^2) = (1/2) / 2
r^2 = 1/4
Taking the square root of both sides:
r = sqrt(1/4) = 1/2
Now, substitute the value of r in equation (1):
a * (1/2)^2 = 2
a * 1/4 = 2
a = 2 * 4
a = 8
Therefore, the first term (a) of the geometric sequence is 8.
3. a. To find the geometric mean (G) between two numbers, you can use the formula:
G = sqrt(a * b)
In this case, let a = 1/2 and b = 1/3:
G = sqrt((1/2) * (1/3))
= sqrt(1/6)
= sqrt(1) / sqrt(6)
= 1 / sqrt(6)
Therefore, the geometric mean between 1/2 and 1/3 is 1 / sqrt(6).
b. To find the geometric mean (G) between two irrational numbers, you can simply take the square root of their product.
In this case, let a = sqrt(5) and b = sqrt(2):
G = sqrt((sqrt(5)) * (sqrt(2)))
= sqrt(sqrt(10))
Therefore, the geometric mean between sqrt(5) and sqrt(2) is sqrt(sqrt(10)).