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Question #40715

Difficulty: 8/10 | 🟢 Free

The Question

Let $n$ arithmetic means $A_1, A_2, \dots, A_n$ be inserted between $4$ and $64$ . Three vectors are defined as follows: $\vec{v}_1 = A_1 \hat{i} + A_2 \hat{j} + A_3 \hat{k}$ $\vec{v}_2 = A_4 \hat{i} + A_5 \hat{j} + A_6 \hat{k}$ $\vec{v}_3 = A_7 \hat{i} + A_8 \hat{j} + (A_9 + \lambda_n) \hat{k}$ where $\lambda_n = \frac{216000}{(n+1)^3} - \frac{43200}{(n+1)^2} + \frac{2820}{n+1} - 60$ . If the vectors $\vec{v}_1, \vec{v}_2, \vec{v}_3$ are linearly dependent, find the sum of all possible values of $n$ .

Options

A.
44
B.
54
C.
33
D.
45

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Insertion of Means|Linear Independence vs Dependence|Elementary Row Operations

Why This Question Is Tricky

Common Mistakes

The problem defines a sequence where the arithmetic means inserted are treated as components of vectors in $\mathbb{R}^3$ .

Students must use Elementary Column Operations on the matrix formed by these vectors to prove their Linear Dependence, which only occurs if the common difference of the sequence satisfies a specific determinant vanishing condition.

The primary trap lies in treating the means purely as scalar sequence elements, whereas the dependency requires a vector space interpretation and matrix analysis.

Furthermore, the problem disguises a cubic polynomial in terms of the common difference $d$ as a rational function of $n$ , forcing candidates to recognize the substitution $d = \frac{60}{n+1}$ to make the determinant tractable.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	Insertion of Means
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free