Question

The quantities \(a_{1},a_{2},a_{3},\ldots\) form an infinite decreasing G.P. If \(a_{1}=1\), then the common ratio of the progression for which the expression \(6a_{5}-16a_{4}-3a_{3}+12a_{2}\) is at a maximum is:

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $-\frac{1}{4}$

Hint:

When factoring high-degree polynomials in optimization questions, always look for grouping opportunities before attempting full synthetic division. Here, noticing that $-8$ is exactly twice $-4$ allows you to pull out the common factor $(r-2)$ instantly!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
An infinite decreasing G.P. implies that the common ratio \( r \) must satisfy \( 0<|r|<1 \). We express the given sum in terms of \( r \) and find its maximum using differentiation.
Step 2: Key Formula or Approach:
1. G.P. terms: \( a_n = a_1 r^{n-1} = r^{n-1} \).
2. Let \( E(r) = 6r^4 - 16r^3 - 3r^2 + 12r \).
3. Solve \( E'(r) = 0 \).
Step 3: Detailed Explanation:
Substitute terms into the expression:
\[ E(r) = 6r^4 - 16r^3 - 3r^2 + 12r^1 \]
Differentiate with respect to \( r \):
\[ E'(r) = 24r^3 - 48r^2 - 6r + 12 \]
To find critical points, set \( E'(r) = 0 \):
\[ 24r^3 - 48r^2 - 6r + 12 = 0 \]
Divide by 6:
\[ 4r^3 - 8r^2 - r + 2 = 0 \]
Factor by grouping:
\[ 4r^2(r - 2) - 1(r - 2) = 0 \]
\[ (4r^2 - 1)(r - 2) = 0 \]
Roots are \( r = 2, r = 1/2, r = -1/2 \).
Since the G.P. is infinite and decreasing, \( |r|<1 \). Thus \( r=2 \) is rejected.
Check the second derivative \( E''(r) = 72r^2 - 96r - 6 \).
At \( r = 1/2 \): \( E''(1/2) = 72(1/4) - 96(1/2) - 6 = 18 - 48 - 6 = -36<0 \) (Maximum).
At \( r = -1/2 \): \( E''(-1/2) = 18 + 48 - 6 = 60>0 \) (Minimum).
Therefore, the expression is maximized when \( r = 1/2 \).
Step 4: Final Answer:
The common ratio is \( 1/2 \), which corresponds to option (B).