The second and sixth terms of a G.P are 7
and 243/18, what is their common ratio?
A. 2.48
B. 3.68
C. 4.98
D. 1.18
Let's denote the second term of the geometric progression as a and the common ratio as r. We know that the second term, a, is equal to 7, and the sixth term is equal to (243/18).
The formula for the nth term of a geometric progression is given by:
Tn = ar^(n-1)
We can set up two equations using the given information:
7 = a * r^(2-1) => 7 = a * r
(243/18) = a * r^(6-1) => (243/18) = a * r^5
Dividing the second equation by the first equation, we can eliminate the term a:
(243/18) / 7 = (a * r^5) / (a * r)
(243/18) / 7 = r^4
(243/18) / 7 = r^(4-1)
(243/18) / 7 = r^3
Simplifying the left-hand side of the equation: 243/18 divided by 7 equals (243/18)*(1/7) = 243/126
243/126 = r^3
Taking the cube root of both sides of the equation, we get:
(r^3)^(1/3) = (243/126)^(1/3)
r = (243/126)^(1/3)
Calculating the cube root of (243/126) using a calculator, we find that r ≈ 1.179535.
Therefore, the common ratio of the geometric progression is approximately 1.18.
The correct answer is D. 1.18.
To find the common ratio of a Geometric Progression (G.P.), we can use the formula:
\[ a_{n} = a \cdot r^{(n-1)} \]
Given that the second term, \( a_{2} \), is 7 and the sixth term, \( a_{6} \), is \(\dfrac{243}{18}\), we have:
\[ a_{2} = a \cdot r^{(2-1)} = 7 \]
\[ a_{6} = a \cdot r^{(6-1)} = \dfrac{243}{18} \]
Let's solve for \( a \) and \( r \):
From the first equation, we have \( 7 = a \cdot r^{(1)} \).
From the second equation, we have \( \dfrac{243}{18} = a \cdot r^{(5)} \).
Divide the second equation by the first equation:
\[ \dfrac{243}{18} \cdot \dfrac{1}{7} = r^{(5-1)} \]
Simplifying the left side:
\[ \dfrac{243}{18 \cdot 7} = r^{4} \]
\[ \dfrac{243}{126} = r^{4} \]
Taking the fourth root of both sides:
\[ r = \sqrt[4]{\dfrac{243}{126}} \]
Evaluating this value gives us:
\[ r \approx 2.48 \approx \text{Option A} \]
Therefore, the common ratio is approximately 2.48.