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)
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)
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!
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).
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).