To find the third term of the exponential sequence, we need to find the common ratio (r) first:
r = (3 - 2√2) / (√2 - 1)
r = (3 - 2√2)(√2 + 1) / (√2 - 1)(√2 + 1)
r = 5 + 2√2
Now we can use the formula for the nth term of a geometric sequence:
an = ar^(n-1)
where a is the first term, r is the common ratio, and n is the term we want to find.
a = √2 - 1
r = 5 + 2√2
n = 3
a3 = (√2 - 1)(5 + 2√2)^(3-1)
a3 = (√2 - 1)(5 + 2√2)^2
a3 = (√2 - 1)(25 + 20√2 + 8)
a3 = (√2 - 1)(33 + 20√2)
a3 = 33√2 + 47 - 20√2 - 14
a3 = 19√2 + 33
Therefore, the third term of the exponential sequence is 19√2 + 33.
. Find the third term of the exponential
sequence (G.P),
(√2 - 1), (3− 2√2), ………
5 answers
Given that n = (
−12
5
) and s = (
1
−1
), Find the
vector q such that |𝒒|= 35 and q is in the
direction of (n + 5s).
−12
5
) and s = (
1
−1
), Find the
vector q such that |𝒒|= 35 and q is in the
direction of (n + 5s).
To find q, we need to first find the vector direction of (n + 5s).
(n + 5s) = (-12, 5) + 5(1, -1) = (-7, 0)
The magnitude of this vector is sqrt((-7)^2 + 0^2) = 7.
We want q to have magnitude 35, so we can scale the direction vector by a factor of 5:
q = 5(n + 5s) = 5(-7, 0) = (-35, 0)
Therefore, the vector q such that |q| = 35 and q is in the direction of (n + 5s) is (-35, 0).
(n + 5s) = (-12, 5) + 5(1, -1) = (-7, 0)
The magnitude of this vector is sqrt((-7)^2 + 0^2) = 7.
We want q to have magnitude 35, so we can scale the direction vector by a factor of 5:
q = 5(n + 5s) = 5(-7, 0) = (-35, 0)
Therefore, the vector q such that |q| = 35 and q is in the direction of (n + 5s) is (-35, 0).
AAAaannndd the bot gets it wrong yet again!
r = (3 - 2√2) / (√2 - 1) = (√2 - 1)
since (√2 - 1)^2 = (3 - 2√2)
so the 3rd term is just (√2 - 1)^3 = 5√2 - 7
r = (3 - 2√2) / (√2 - 1) = (√2 - 1)
since (√2 - 1)^2 = (3 - 2√2)
so the 3rd term is just (√2 - 1)^3 = 5√2 - 7
I apologize for the mistake in my previous response. You are correct that the common ratio is indeed (√2 - 1) and not (5 + 2√2).
Using the formula for the nth term of a geometric sequence, we have:
a = √2 - 1
r = (√2 - 1)
n = 3
a3 = (√2 - 1)(√2 - 1)^(3-1)
a3 = (√2 - 1)(√2 - 1)^2
a3 = (√2 - 1)(3 - 2√2)
a3 = 3√2 - 5 - 2√2 + 2
a3 = 5√2 - 3
Therefore, the third term of the exponential sequence is 5√2 - 3.
Using the formula for the nth term of a geometric sequence, we have:
a = √2 - 1
r = (√2 - 1)
n = 3
a3 = (√2 - 1)(√2 - 1)^(3-1)
a3 = (√2 - 1)(√2 - 1)^2
a3 = (√2 - 1)(3 - 2√2)
a3 = 3√2 - 5 - 2√2 + 2
a3 = 5√2 - 3
Therefore, the third term of the exponential sequence is 5√2 - 3.
5 answers