Given geometric sequence tn find t6 and S∞ if t2 = 243 and t5=9
Well well, it seems like we have a geometric sequence puzzle to solve! Let's get down to business.
In a geometric sequence, each term is found by multiplying the previous term by a common ratio (let's call it "r"). Since we know t2 is 243 and t5 is 9, we can use this information to figure out r.
To find r, we divide t5 by t2:
r = t5 / t2 = 9 / 243 = 1/27
Now that we have our common ratio, we can find t6 by multiplying t5 by r:
t6 = t5 * r = 9 * (1/27) = 1/3
Next up, let's calculate the sum to infinity (S∞) of this geometric sequence. To do that, we can use the formula:
S∞ = t1 / (1 - r)
Since we know t1 equals t2 multiplied by r, we have:
S∞ = t2 * r / (1 - r)
Now, plugging in the values:
S∞ = 243 * (1/27) / (1 - 1/27)
Simplifying further, we get:
S∞ = 3/26
So, t6 is 1/3, and the sum to infinity (S∞) is 3/26. Isn't math just a barrel of laughs?
To find t6 and S∞ of a geometric sequence, we need to find the common ratio (r) of the sequence first.
Given that t2 = 243 and t5 = 9, we can find the common ratio (r) using the formula:
tₙ = t₁ * r^(n-1)
Let's solve for r using the given information.
For t2 = 243:
243 = t₁ * r^(2-1)
243 = t₁ * r
For t5 = 9:
9 = t₁ * r^(5-1)
9 = t₁ * r^4
Now, we can solve these two equations simultaneously. Divide the second equation by the first equation to eliminate t₁:
(9 / 243) = (t₁ * r^4) / (t₁ * r)
0.037 = r^3
Taking the cube root of both sides gives us:
r ≈ 0.302
Now that we have the common ratio (r), we can find t6 and S∞.
To find t6, we use the formula:
t₆ = t₁ * r^(6-1)
Substituting the value of r, we have:
t₆ = t₁ * (0.302)^(5)
To find S∞ (the sum of an infinite geometric sequence), we use the formula:
S∞ = t₁ / (1 - r)
Substituting the value of r, we have:
S∞ = t₁ / (1 - 0.302)
Please provide the value of t₁ (the first term) to proceed with the calculations.
To find the sixth term (t6) and the sum to infinity (S∞) of a geometric sequence, we first need to determine the common ratio (r).
In a geometric sequence, each term is obtained by multiplying the previous term by the common ratio. We can use this property to setup the following equations using the given information:
t2 = t1 * r (Equation 1)
t5 = t1 * r^4 (Equation 2)
Let's solve for t1 and r using these equations.
From Equation 1, we have:
t2 = t1 * r
243 = t1 * r
From Equation 2, we have:
t5 = t1 * r^4
9 = t1 * r^4
Now we can set up a ratio by dividing the second equation by the first equation:
9/243 = (t1 * r^4) / (t1 * r)
1/27 = r^3
We can rewrite this as:
r^3 = 1/27
Now we calculate the cube root of both sides to find the value of r:
r = ∛(1/27)
r = 1/3
Now that we know the value of r, we can substitute it back into one of the original equations to find t1:
243 = t1 * (1/3)^2
243 = t1 * (1/9)
t1 = 243 * 9
t1 = 2187
Now we have t1 = 2187 and r = 1/3.
To find t6, we use the formula for the nth term of a geometric sequence:
tn = t1 * r^(n-1)
t6 = 2187 * (1/3)^(6-1)
t6 = 2187 * (1/3)^5
t6 = 2187 * (1/243)
t6 = 9
Therefore, the sixth term (t6) is 9.
To find the sum to infinity (S∞), we can use the formula for the sum of an infinite geometric sequence:
S∞ = t1 / (1 - r)
S∞ = 2187 / (1 - 1/3)
S∞ = 2187 / (2/3)
S∞ = 2187 * (3/2)
S∞ = 3280.5
Therefore, the sum to infinity (S∞) is approximately 3280.5.
the fact that S∞ exists means that | r | < 1
... two terms that are three apart (t2 and t5) are both positive, so r is positive
243 r^3 = 9 ... r = ?
find r and t1 and plug them into the sum equation