Question

The sum of the second and third terms of a G.P. is 8 and the fourth term is 4. The common ratio \(r \neq 1\) is

• A. \(-\frac{1}{2}\)
• B. \(\frac{1}{2}\)
• C. \(-\frac{1}{4}\)
• D. \(-\frac{1}{3}\)
• E. \(\frac{1}{4}\)

Hint:

In G.P. problems, dividing one equation by another is the standard trick to eliminate the first term \(a\). Always look for algebraic ways to reduce the power of \(r\) during division.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Let the terms of the G.P. be \(a, ar, ar^2, ar^3, \dots\).
We translate the given text into mathematical equations and solve for \(r\).

Step 2: Key Formula or Approach:
1. \(T_2 + T_3 = 8 \implies ar + ar^2 = 8\).
2. \(T_4 = 4 \implies ar^3 = 4\).

Step 3: Detailed Explanation:
From equation 1:
\[ ar(1 + r) = 8 \dots (i) \]
From equation 2:
\[ ar^3 = 4 \dots (ii) \]
Divide equation (i) by equation (ii) to eliminate \(a\):
\[ \frac{ar(1 + r)}{ar^3} = \frac{8}{4} \]
\[ \frac{1 + r}{r^2} = 2 \]
Cross-multiply:
\[ 1 + r = 2r^2 \]
Rearrange into a standard quadratic equation:
\[ 2r^2 - r - 1 = 0 \]
Factoring the quadratic:
\[ 2r^2 - 2r + r - 1 = 0 \]
\[ 2r(r - 1) + 1(r - 1) = 0 \]
\[ (2r + 1)(r - 1) = 0 \]
Possible values for \(r\) are \(1\) or \(-\frac{1}{2}\).
Since the question states \(r \neq 1\), the only valid solution is \(r = -\frac{1}{2}\).

Step 4: Final Answer:
The common ratio is \(-\frac{1}{2}\).