Question

The sum of the first two terms of a geometric series is 12 and the third term is 16. Then the common ratio $r > 0$ of the geometric progression is

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. 2
• D. 3
• E. $\frac{3}{2}$

Hint:

Always check constraints like $r > 0$ early to eliminate impossible roots from quadratic equations immediately.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In a Geometric Progression (GP), the terms are expressed as \( a, ar, ar^2, \dots \). We use the given conditions to set up equations for the first term \( a \) and the common ratio \( r \).

Step 2: Detailed Explanation:
Let the first term be $a$ and the common ratio be $r$.
Condition 1: Sum of the first two terms is 12.
\[ a + ar = 12 \implies a(1 + r) = 12 \quad \dots \text{(Eqn 1)} \]
Condition 2: The third term is 16.
\[ ar^2 = 16 \implies a = \frac{16}{r^2} \quad \dots \text{(Eqn 2)} \]
Substitute Eqn 2 into Eqn 1:
\[ \frac{16}{r^2}(1 + r) = 12 \]
Divide both sides by 4:
\[ \frac{4}{r^2}(1 + r) = 3 \]
\[ 4 + 4r = 3r^2 \]
\[ 3r^2 - 4r - 4 = 0 \]
Solve using the quadratic formula or factoring:
\[ 3r^2 - 6r + 2r - 4 = 0 \]
\[ 3r(r - 2) + 2(r - 2) = 0 \]
\[ (3r + 2)(r - 2) = 0 \]
This gives \( r = 2 \) or \( r = -2/3 \).
Since the question specifies \( r > 0 \), the common ratio must be 2.

Step 3: Final Answer:
The common ratio is 2.