the third term of a geometric sequence is 6 and the sixth term is 3 over 32. Determine the first terms and the general term Tn?
Oh, so we have a geometric sequence going on here? Great! Let's put on our funny glasses and solve it.
To find the first term, let's call it "a," we can use the formula for the nth term of a geometric sequence: Tn = a * r^(n-1).
Now, we know that the third term (T3) is 6. Let's plug that in and get a * r^2 = 6.
Moving on to the sixth term (T6), which is 3/32, we can substitute it into the formula: a * r^(6-1) = 3/32.
Okay, let's do some math magic here. Since we have two equations with a and r, we can solve them simultaneously and find their values.
Dividing the second equation by the first, we get (a * r^5) / (a * r^2) = (3/32) / 6.
Simplifying, we find r^3 = 1/64.
Taking the cube root of both sides, we get r = 1/4.
Now, to find a, we can substitute r into the first equation.
So, a * (1/4)^2 = 6.
Simplifying, we get a/16 = 6.
Therefore, a = 96.
So, the first term is 96, and the general term Tn can be written as Tn = 96 * (1/4)^(n-1). And there you have it!
Now wasn't that an amusing way to solve it? I hope I put a smile on your face!
To find the first term and the general term of a geometric sequence, we can use the formula:
Tn = a * r^(n-1)
where Tn represents the nth term, a is the first term, r is the common ratio, and n is the position of the term.
Given the information that the third term is 6 and the sixth term is 3/32, we can set up the following equations:
For the third term:
T3 = a * r^(3-1) = 6 .....(1)
For the sixth term:
T6 = a * r^(6-1) = 3/32 .....(2)
Simplifying equation (1), we have:
a * r^2 = 6
And simplifying equation (2), we have:
a * r^5 = 3/32
Now, we can solve this system of equations to find the values of a and r.
Dividing equation (2) by equation (1), we get:
(a * r^5) / (a * r^2) = (3/32) / 6
Simplifying this further, we have:
r^(5-2) = (3/32) / 6
r^3 = (3/32) / 6
Next, simplify the right side of the equation:
r^3 = 1/64
Taking the cube root of both sides, we find:
r = (1/64)^(1/3)
Simplifying this, we get:
r = 1/4
Now, substituting the value of r into equation (1), we can solve for a:
a * (1/4)^2 = 6
Multiplying both sides by 16 (to eliminate the fraction), we have:
4a = 96
Dividing both sides by 4, we find:
a = 24
Therefore, the first term (a) is 24 and the common ratio (r) is 1/4.
Now, we can write the general term (Tn) as:
Tn = 24 * (1/4)^(n-1)
So, the general term is Tn = 24 * (1/4)^(n-1).
To find the first term (a) and the general term (Tn) of a geometric sequence, we can use the following formulas:
a = T1 = first term
Tn = a * r^(n-1) ----(1)
where:
r = common ratio
n = term number
Given that the third term is 6, we can use equation (1) to find the common ratio (r). Substituting n = 3 and T3 = 6 into equation (1):
6 = a * r^(3-1)
6 = a * r^2
Similarly, for the sixth term being 3/32:
3/32 = a * r^(6-1)
3/32 = a * r^5
Now, we have a system of equations:
6 = a * r^2 ----(2)
3/32 = a * r^5 ----(3)
To solve this system of equations, we need to eliminate one of the variables (either a or r). Let's eliminate a:
Divide equation (2) by equation (3):
(6)/(3/32) = (a * r^2) / (a * r^5)
(6) * (32/3) = r^2 / r^5
64 = r^(-3) (using the exponent rule r^a / r^b = r^(a-b))
Taking the cube root of both sides:
∛(64) = ∛(r^(-3))
4 = r^(-1) (∛(64) = 4)
Taking the reciprocal of both sides:
1/4 = r
Substituting this value back into equation (2) to solve for 'a':
6 = a * (1/4)^2
6 = a * 1/16
96 = a
Therefore, the first term (a) is 96 and the common ratio (r) is 1/4.
Now that we know the values of 'a' and 'r', we can write the general term (Tn) using equation (1):
Tn = a * r^(n-1)
Tn = 96 * (1/4)^(n-1)
So, the general term (Tn) is 96 * (1/4)^(n-1).
1 a
2 ar
3 ar^2
4 ar^3
5 ar^4
6 ar^5
n ar^(n-1)
so
ar^2 = 6
ar^5 = 3/32
so
a = 6/r^2
(6/r^2)r^5 = 3/32
6/r^3 = 3/32
r^3 = 32*2 = 64
r = 4
then
a = 6/16 = 3/4
so
Tn = (3/4)4^(n-1)