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:
A geometric progression has terms \(a, ar, ar^2, \dots\).
We have two conditions forming a system of algebraic equations in terms of \(a\) and \(r\).
Step 2: Key Formula or Approach:
The sum of the first two terms: \(a + ar = 12\).
The third term: \(ar^2 = 16\).
Express \(a\) in terms of \(r\) and substitute into the other equation.
Step 3: Detailed Explanation:
From \(ar^2 = 16\), we get \(a = \frac{16}{r^2}\).
Substitute this into \(a(1 + r) = 12\):
\[ \left(\frac{16}{r^2}\right)(1 + r) = 12 \] Divide by 4:
\[ \frac{4(1 + r)}{r^2} = 3 \] Cross-multiply to form a quadratic equation:
\[ 4 + 4r = 3r^2 \implies 3r^2 - 4r - 4 = 0 \] Factorize the quadratic:
\[ 3r^2 - 6r + 2r - 4 = 0 \] \[ 3r(r - 2) + 2(r - 2) = 0 \implies (3r + 2)(r - 2) = 0 \] The roots are \(r = -\frac{2}{3}\) and \(r = 2\).
Since the condition specifies \(r>0\), we choose \(r = 2\).
Step 4: Final Answer:
The common ratio is 2.