The six terms of a G. P is 20 and the 12th term is 160. Find the common ratio and first term
To find the common ratio and first term of a geometric progression (G.P.), we can use the formulas:
For the nth term of a G.P.:
\[a_n = a_1 \cdot r^{(n-1)}\]
(Note: \(a_n\) represents the nth term, \(a_1\) represents the first term, and \(r\) represents the common ratio)
We are given that the 6th term (\(a_6\)) is equal to 20, so we can substitute those values into the formula:
\[20 = a_1 \cdot r^{(6-1)}\]
Simplifying the exponent:
\[20 = a_1 \cdot r^5 \quad (Equation 1)\]
We are also given that the 12th term (\(a_{12}\)) is equal to 160, so we can substitute those values into the formula:
\[160 = a_1 \cdot r^{(12-1)}\]
Simplifying the exponent:
\[160 = a_1 \cdot r^{11} \quad (Equation 2)\]
Now we have a system of two equations (Equation 1 and Equation 2) with two unknowns (\(a_1\) and \(r\)). To solve this system, we can divide Equation 2 by Equation 1:
\[\frac{160}{20} = \frac{a_1 \cdot r^{11}}{a_1 \cdot r^5}\]
Simplifying:
\[8 = r^6\]
To find the value of \(r\), we can take the 6th root of both sides:
\[\sqrt[6]{8} = r\]
\[r = 2\]
Now that we have the value of \(r\), we can substitute it into Equation 1 to find the value of \(a_1\):
\[20 = a_1 \cdot 2^5\]
Simplifying:
\[20 = 32a_1\]
Dividing both sides by 32:
\[\frac{20}{32} = a_1\]
\[a_1 = \frac{5}{8}\]
Therefore, the common ratio (\(r\)) is 2, and the first term (\(a_1\)) is \(\frac{5}{8}\).
To find the common ratio and first term of a geometric progression (G.P.), we can use the formula:
\[a_n = a_1 \cdot r^{(n-1)}\]
where:
- \(a_n\) is the \(n\)th term of the G.P.
- \(a_1\) is the first term of the G.P.
- \(r\) is the common ratio.
- \(n\) is the position of the term in the G.P.
Given that the 6th term is 20 and the 12th term is 160, we can set up two equations using the formula.
Equation 1 (using the 6th term):
\[20 = a_1 \cdot r^{(6-1)}\]
Equation 2 (using the 12th term):
\[160 = a_1 \cdot r^{(12-1)}\]
Now we have a system of two equations with two unknowns (the first term, \(a_1\), and the common ratio, \(r\)).
To solve these equations, we can first simplify them:
Equation 1:
\[20 = a_1 \cdot r^5\]
Equation 2:
\[160 = a_1 \cdot r^{11}\]
Next, divide Equation 2 by Equation 1:
\[\frac{160}{20} = \frac{a_1 \cdot r^{11}}{a_1 \cdot r^5}\]
Simplifying, we get:
\[8 = r^6\]
Now, take the 6th root of both sides to isolate \(r\):
\[\sqrt[6]{8} = \sqrt[6]{r^6}\]
Simplifying further, we have:
\[2 = r\]
Therefore, the common ratio (\(r\)) is 2.
To find the first term (\(a_1\)), we can substitute the value of \(r\) into one of the original equations. Let's use Equation 1:
\[20 = a_1 \cdot 2^5\]
Simplifying further:
\[20 = a_1 \cdot 32\]
Now, divide both sides by 32 to solve for \(a_1\):
\[\frac{20}{32} = \frac{a_1 \cdot 32}{32}\]
Simplifying:
\[\frac{5}{8} = a_1\]
Therefore, the first term (\(a_1\)) is \(5/8\).