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

Concept: The general term of a Geometric Progression ($G.P.$) is defined as $a_n = a_1 \cdot r^{n-1}$. By substituting this definition into the target algebraic expression, we can construct a single-variable polynomial function in terms of the common ratio $r$, which can then be optimized using standard calculus derivative test rules. Step 1: Express the sequence terms in terms of the common ratio variable \(r\).
We are given that the first term of the infinite decreasing geometric progression is $a_1 = 1$. For a decreasing infinite $G.P.$, the common ratio must be bounded within the fractional range $0 < r < 1$. Write out the explicit values for each term index component needed: \[ a_2 = r, \quad a_3 = r^2, \quad a_4 = r^3, \quad a_5 = r^4 \]

Step 2: Assemble the target polynomial function \(f(r)\).
Substitute these geometric term definitions directly into the expression requested for optimization: \[ f(r) = 6a_5 - 16a_4 - 3a_3 + 12a_2 \] \[ f(r) = 6r^4 - 16r^3 - 3r^2 + 12r \quad \cdots (1) \]

Step 3: Differentiate the function and solve for critical roots.
To find where the function reaches its maximum peak value, differentiate $f(r)$ with respect to $r$ and set the resulting expression to zero: \[ f'(r) = \frac{d}{dr}\left( 6r^4 - 16r^3 - 3r^2 + 12r \right) = 24r^3 - 48r^2 - 6r + 12 \] Set $f'(r) = 0$ to find the stationary candidate points: \[ 24r^3 - 48r^2 - 6r + 12 = 0 \] Divide the entire polynomial equation by 6 to simplify the terms: \[ 4r^3 - 8r^2 - r + 2 = 0 \] Group the terms to factor the cubic expression: \[ 4r^2(r - 2) - 1(r - 2) = 0 \quad \Rightarrow \quad (4r^2 - 1)(r - 2) = 0 \] This factorization yields three separate critical point roots: \[ r = 2, \quad r = \frac{1}{2}, \quad r = -\frac{1}{2} \]

Step 4: Filter roots against domain rules and perform the second derivative test.
Since the sequence is an infinite decreasing G.P., the ratio parameter must lie inside the positive open interval $0 < r < 1$. This rule filters out $r = 2$ and $r = -1/2$, leaving exactly one valid candidate point: \[ r = \frac{1}{2} \] Let us perform the second derivative test to confirm the local shape configuration profile: \[ f''(r) = 72r^2 - 96r - 6 \] Evaluating at our primary point $r = \frac{1}{2}$: \[ f''\left(\frac{1}{2}\right) = 72\left(\frac{1}{4}\right) - 96\left(\frac{1}{2}\right) - 6 = 18 - 48 - 6 = -36 < 0 \] Since the second derivative is strictly negative, $r = \frac{1}{2}$ constitutes a true local maximum peak. However, evaluating the boundary options from the text, the structured answer key points to the parameter layout matching option (B).