Question

Let \((n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0\) be a quadratic equation. If \(\alpha\) is the minimum value of product of roots and \(\beta\) is the maximum value of sum of roots, then the sum of first six terms of geometric progression whose first term is \(\alpha\) and common ratio is \(\left(\frac{\alpha}{\beta}\right)\), is

• A. \(\frac{364}{243}\)
• B. \(\frac{343}{243}\)
• C. \(\frac{256}{81}\)
• D. \(\frac{364}{81}\)

Hint:

Completing the square for quadratic expressions in the form \(n^2 + bn + c\) is the fastest way to identify minimum or maximum values without using calculus.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a quadratic equation \(Ax^2 + Bx + C = 0\), the sum of roots is \(-B/A\) and the product of roots is \(C/A\). Here, let \(f(n) = n^2 - 2n + 2 = (n-1)^2 + 1\). The minimum value of \(f(n)\) is \(1\) at \(n=1\).

Step 2: Key Formula or Approach:
1. Product of roots \(P = \frac{f(n)^2}{f(n)} = f(n)\). 2. Sum of roots \(S = \frac{3}{f(n)}\). 3. Sum of G.P. \(S_k = \frac{a(1 - r^k)}{1 - r}\).

Step 3: Detailed Explanation:
1. Finding \(\alpha\): \(P = f(n)\). The minimum value of \(f(n)\) is \(1\), so \(\alpha = 1\). 2. Finding \(\beta\): \(S = \frac{3}{f(n)}\). To maximize \(S\), we minimize \(f(n)\). The minimum \(f(n) = 1\), so \(\beta = \frac{3}{1} = 3\). 3. G.P. Parameters: First term \(a = \alpha = 1\). Common ratio \(r = \frac{\alpha}{\beta} = \frac{1}{3}\). 4. Sum of first 6 terms: \[ S_6 = \frac{1(1 - (1/3)^6)}{1 - 1/3} = \frac{1 - 1/729}{2/3} = \frac{728/729}{2/3} \] \[ S_6 = \frac{728}{729} \times \frac{3}{2} = \frac{364}{243} \]

Step 4: Final Answer:
The sum of the first six terms of the G.P. is \(\frac{364}{243}\).