Which term of the Geometric sequence with first two terms 192 and 96 is equal to 3
a = 192
r = 96/192 = 1/2
term(n) = a r^(n-1)
3 = 192 (1/2)^(n-1)
1/64 = (1/2)^(n-1)
(1/2)^6 = (1/2)^(n-1)
n-1 = 6
n = 7
It is the 7th term
check:
term(7) = a r^6
= 192 (1/2)^6 = 192(1/64) = 3
To find which term of the geometric sequence with first two terms 192 and 96 is equal to 3, we can use the formula for the nth term of a geometric sequence:
\(a_n = a_1 \times r^{(n-1)}\)
where:
\(a_n\) is the nth term of the sequence,
\(a_1\) is the first term of the sequence,
\(r\) is the common ratio, and
\(n\) is the term number we're looking for.
In this case, we know that \(a_1 = 192\) and \(a_2 = 96\). We need to find the value of \(n\) when \(a_n = 3\).
Substitute the given values into the formula and solve for \(n\):
\(3 = 192 \times r^{(n-1)}\)
Since we know that \(a_2 = 96\), we can rewrite \(a_2\) using the formula:
\(96 = 192 \times r^{(2-1)}\)
Simplify:
\(96 = 192 \times r^1\)
\(r = \frac{96}{192}\)
\(r = \frac{1}{2}\)
Now substitute the value of \(r\) into the equation for \(a_n\) using \(a_1 = 192\) and solve for \(n\):
\(3 = 192 \times \left(\frac{1}{2}\right)^{(n-1)}\)
Divide both sides of the equation by 192:
\(\frac{3}{192} = \left(\frac{1}{2}\right)^{(n-1)}\)
Simplify:
\(\frac{1}{64} = \left(\frac{1}{2}\right)^{(n-1)}\)
To solve for \(n\), take the logarithm of both sides of the equation. Let's use the base-2 logarithm:
\(\log_2\left(\frac{1}{64}\right) = \log_2\left(\left(\frac{1}{2}\right)^{(n-1)}\right)\)
Simplify:
\(\log_2\left(2^{-6}\right) = (n-1) \cdot \log_2\left(\frac{1}{2}\right)\)
Using the properties of logarithms, we can simplify further:
\(-6 \cdot \log_2(2) = (n-1) \cdot (-1)\)
\(-6 = n - 1\)
Add 1 to both sides of the equation:
\(n = -6 + 1\)
\(n = -5\)
The term number \(n\) is -5, which means that there is no term in the geometric sequence that is equal to 3.
To find which term of the geometric sequence is equal to 3, we can use the formula for the nth term of a geometric sequence:
\(a_n = a_1 \times r^{(n-1)}\)
Where:
\(a_n\) is the nth term of the sequence,
\(a_1\) is the first term of the sequence,
r is the common ratio,
and n is the term number we want to find.
Given that the first two terms of the geometric sequence are 192 and 96, we can substitute them into the formula as follows:
For the first term (n = 1):
\(192 = a_1 \times r^{(1-1)}\)
\(192 = a_1\)
For the second term (n = 2):
\(96 = a_1 \times r^{(2-1)}\)
\(96 = a_1 \times r\)
Now, we can solve the equations to find the values of \(a_1\) and r:
Substituting \(192\) for \(a_1\) in the second equation:
\(96 = 192 \times r\)
\(r = \frac{96}{192}\)
\(r = \frac{1}{2}\)
Now that we have found the common ratio (r), we can substitute it into the first equation to find the value of \(a_1\):
\(192 = a_1\)
Therefore, \(a_1 = 192\) and \(r = \frac{1}{2}\).
To find the term number that is equal to 3, we can substitute the values of \(a_1\) and r into the formula and solve for n:
\(3 = 192 \times \left(\frac{1}{2}\right)^{(n-1)}\)
To solve for n, we can take the logarithm of both sides:
\(log(3) = log\left(192 \times \left(\frac{1}{2}\right)^{(n-1)}\right)\)
Using logarithmic properties, we can rewrite the equation as:
\(log(3) = log(192) + (n-1) \times log\left(\frac{1}{2}\right)\)
Now, we can isolate n:
\(log(3) - log(192) = (n-1) \times log\left(\frac{1}{2}\right)\)
\((n-1) \times log\left(\frac{1}{2}\right) = log(192) - log(3)\)
Dividing both sides by log(1/2):
\(n-1 = \frac{log(192) - log(3)}{log\left(\frac{1}{2}\right)}\)
Adding 1 to both sides:
\(n = \frac{log(192) - log(3)}{log\left(\frac{1}{2}\right)} + 1\)
Using a calculator, we can now evaluate the right side of the equation:
\(n \approx 7.584\)
Therefore, the term of the geometric sequence that is equal to 3 is approximately the 8th term (since n represents the term number).