The six terms of a G. P is 20 and the 12th term is 160. Find the common ratio
Let the first term of the G. P be a and the common ratio be r.
We know that the sixth term is 20, so we can write:
a * r^5 = 20 ......(1)
We also know that the twelfth term is 160, so we can write:
a * r^11 = 160 ......(2)
To find the value of r, we divide equation (2) by equation (1):
(r^11) / (r^5) = 160 / 20
r^6 = 8
Taking the sixth root on both sides:
r = ∛(8) = 2
Therefore, the common ratio is 2.
To find the common ratio (r) of a geometric progression (G.P.), we can use the formula:
nth term = a * r^(n-1)
where,
nth term is the term at the given position,
a is the first term of the G.P.,
r is the common ratio,
n is the position of the term.
Given that the first term (a) is 20 and the 12th term (nth term) is 160, we can substitute the values into the formula as follows:
160 = 20 * r^(12-1)
Simplifying the equation:
160 = 20 * r^11
Divide both sides of the equation by 20:
8 = r^11
Taking the 11th root of both sides:
r = √[11th root of 8]
Calculating √[11th root of 8]:
r ≈ 1.4678
Therefore, the common ratio (r) of the G.P. is approximately 1.4678.
To find the common ratio (r) of a geometric progression (G.P.), we can use the formula:
\[a_n = a_1 \times r^{(n-1)}\]
Where:
- \(a_n\) is the nth term of the G.P.
- \(a_1\) is the first term of the G.P.
- \(r\) is the common ratio of the G.P.
- \(n\) is the position of the term in the G.P.
In this case, we need to find the common ratio (r). We are given that the first term (\(a_1\)) is 20 and the 12th term (\(a_{12}\)) is 160.
We can start by substituting the given values into the formula:
\[a_{12} = a_1 \times r^{(12-1)}\]
Plugging in the known values:
\[160 = 20 \times r^{11}\]
Next, we can solve for r by isolating it:
\[r^{11} = \frac{160}{20} = 8\]
Taking the 11th root of both sides:
\[r = \sqrt[11]{8}\]
Calculating:
\[r ≈ 1.2649\]
Therefore, the common ratio (r) of the geometric progression is approximately 1.2649.