The fourth term of a GP is 2 and the a sixth term is 8. Find the common ratio and first term?
ar³=2......(1)
ar^5=8.....(2)
r^5/r³=8/2
r^(5-3)=4
r^2=4
r=2
Find a
Why did the geometric sequence join the circus? Because it wanted to find its common "ratio"!
Now, let's get to business. We have the fourth term, which is 2, and the sixth term, which is 8. To find the common ratio (r) and the first term (a), we can use the formula to calculate any term in a geometric sequence:
Tn = a * r^(n-1),
where Tn is the nth term, a is the first term, r is the common ratio, and n is the term number.
Now let's solve this puzzle!
For the fourth term, we have:
2 = a * r^(4-1).
And for the sixth term, we have:
8 = a * r^(6-1).
Let's divide the two equations to get rid of 'a':
(8/2) = (a * r^(4-1)) / (a * r^(6-1)).
Simplifying that, we have:
4 = (r^3) / (r^5).
To simplify even further, we subtract the exponents of r:
4 = (r^3) / (r^5) = r^(3-5) = 1/r^2.
Now it's time to solve for r!
Cross multiplying, we have:
4 * r^2 = 1.
Dividing both sides by 4 gives us:
r^2 = 1/4.
Taking the square root, we find that:
r = ±√(1/4) = ±(1/2).
Since we're looking for a real number common ratio, let's choose the positive value of 1/2:
r = 1/2.
Now that we've found the common ratio, we can substitute it back into one of the equations to solve for the first term, a. Let's use the equation with the fourth term:
2 = a * (1/2)^(4-1).
Simplifying further, we have:
2 = a * (1/2)^3 = a * (1/8) = a/8.
Now, we can solve for 'a':
2 * 8 = a,
16 = a.
Thus, the common ratio (r) is 1/2, and the first term (a) is 16.
Keep those math jokes coming!
To find the common ratio and first term of a geometric progression (GP), we can use the property that the ratio between consecutive terms in a GP is constant.
Let's denote the first term of the GP as 'a' and the common ratio as 'r'.
According to the given information:
The fourth term of the GP is 2, so we have: a * r^3 = 2 --(1)
The sixth term of the GP is 8, so we have: a * r^5 = 8 --(2)
Now, let's solve these two equations to find the values of 'a' and 'r'.
Divide equation (2) by equation (1) to eliminate 'a':
(a * r^5) / (a * r^3) = 8 / 2
r^2 = 4
Taking the square root of both sides, we get:
r = 2 or r = -2
We have two possible values for 'r', but since the terms of a GP cannot be negative, we can discard r = -2.
So, the common ratio, r, is 2.
Substituting this value of 'r' into equation (1):
a * 2^3 = 2
a * 8 = 2
a = 2/8
a = 1/4
Therefore, the common ratio is 2 and the first term is 1/4.
To find the common ratio and first term of a geometric progression (GP) from the given information, we can use the formula for the nth term of a GP:
\[a_n = a \cdot r^{(n-1)}\]
where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the position of the term.
Given that the fourth term (\(a_4\)) is 2, we can substitute these values into the formula as follows:
\[2 = a \cdot r^{(4-1)}\]
\[2 = a \cdot r^3\]
Similarly, for the sixth term (\(a_6\)) being 8, we can write:
\[8 = a \cdot r^{(6-1)}\]
\[8 = a \cdot r^5\]
Now we have a system of two equations:
\[2 = a \cdot r^3\] (equation 1)
\[8 = a \cdot r^5\] (equation 2)
To determine the values of \(a\) and \(r\), we can divide equation 2 by equation 1:
\[\frac{8}{2} = \frac{a \cdot r^5}{a \cdot r^3}\]
\[4 = r^2\]
Taking the square root of both sides, we get:
\[r = 2\]
Substituting this value of \(r\) into equation 1, we can solve for \(a\):
\[2 = a \cdot (2^3)\]
\[2 = 8a\]
\[a = \frac{2}{8} = \frac{1}{4}\]
Therefore, the common ratio (\(r\)) is 2 and the first term (\(a\)) is \(\frac{1}{4}\).