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Explanation:

Given,

x, y, z are in GP

Thus, 

⇒ y² =xz

⇒ \(2lny = lnx + lnz\)

lnx, ln y and lnz are in AP

⇒ (1 + lnx), (1 + lny) and (1 + lnz) are in AP

⇒ \(\frac{1}{lnx}, \frac{1}{lny}, \frac{1}{lnz}\) are in HP

Concept:

Determinant and Column Operations:

• The determinant of a matrix is a scalar value representing area, volume, or invertibility.
• Column operations like \( C_1 \rightarrow C_1 + C_2 + C_3 \) can simplify a determinant.
• If one column (or row) becomes a linear combination of others or zero, the determinant becomes 0.
• \( \omega \) is a cube root of unity with identity: \( 1 + \omega + \omega^2 = 0 \).

Calculation:

Let \( I = \begin{bmatrix} 1+\omega & 1+\omega^2 & \omega + \omega^2 \ 1 & \omega & \omega^2 \ \frac{1}{\omega} & \frac{1}{\omega^2} & 1 \end{bmatrix} \)

⇒ Apply column operation: \( C_1 \rightarrow C_1 + C_2 + C_3 \)

⇒ First column becomes:

\( C_1 = \begin{bmatrix} (1+\omega) + (1+\omega^2) + (\omega + \omega^2) \ 1 + \omega + \omega^2 \ \frac{1}{\omega} + \frac{1}{\omega^2} + 1 \end{bmatrix} \)

⇒ Using identity: \( 1 + \omega + \omega^2 = 0 \)

⇒ First column becomes: \( \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \)

⇒ New matrix:

\( I = \begin{bmatrix} 0 & 1+\omega^2 & \omega + \omega^2 \ 0 & \omega & \omega^2 \ 0 & \frac{1}{\omega^2} & 1 \end{bmatrix} \)

⇒ Since first column is all zero,

∴ Determinant I = 0

Explanation:

Given

S_n = n(2n + 1), S_n–1 = (n – 1)(2n – 1) 

n^th term = S_n – S_{n – 1}

= n(2n + 1) – (n – 1)(2n–1)

=  4n – 1

∴ Option (a) is correct.

Explanation:

Word 'INDIA' contains three vowels (I, I and A) and two consonants (N and D).

Three vowels (I, I and A) can be arranged at odd position in \(\frac{3!}{2!} = 3 \) ways

2 consonants(N and D) can be arranged in 2! ways = 2 ways. 

Thus, required numbers of ways = (3) (2) = 6.

∴ Option (b) is correct.

Concept:

Permutations and Arrangements:

• A permutation is an arrangement of objects in a specific order. The number of ways to arrange 'n' objects is given by the factorial of 'n' denoted by n!.
• For a set of distinct vowels and consonants, the total number of ways to arrange them can be calculated by multiplying the factorials of the number of vowels and consonants separately.

Calculation:

Given the word 'EQUATION', we have:

• 5 vowels: A, E, I, O, U
• 3 consonants: Q, T, N

⇒ Number of ways = (3!) . (5!)

⇒ Numbers of ways = (3!) . (5!)

⇒ Total required number of ways = (3!) . (5!) + (3!) . (5!)

= 1440

∴ Option (c) is correct.

Calculation:

n is a root of the equation x2 + px + m = 0

So, n² + pn + n = 0

 ..... (1)

Also, m is the root of the equation of x² + px + n = 0

So, m² + pm + n = 0 ..... (2)

Now Eq. (i) – Eq. (ii), we get

⇒ \((n^2 – m^2) + p(n – m) – (n – m) = 0 \)

⇒ \((n – m) [(n + m) + p – 1] = 0 \)

⇒ \(n + m + p – 1 = 0\) ..... ( n ≠ m)

⇒ n + m + p = 1

∴ Option (3) is correct.

Explanation:

Given,

Number of courses, n > 4

The student needs to choose (n - 2) courses out of n courses, with 2 courses being compulsory

∴Total courses is n – 2 and total remaining courses that can be chosen from (n – 2) courses is ( n – 4).

∴ Required number of ways ^(n-2) C _(n-4).

⇒ \(\frac{(n-2)(n-3)}{2}\)

∴ Option d is correct.

Concept:

Determinant of a 3×3 Matrix:

• The determinant of a 3×3 matrix can be found by the cofactor expansion method.
• It is expressed as: \( \text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg) \)
• If the matrix has a common factor across rows or columns, it can be factored out to simplify calculations.
• \( \sum_{n=1}^{k} f(n) \) is a summation notation indicating summing function values from n = 1 to n = k.
• Important identities used:
• \( \sum_{n=1}^{k} n = \frac{k(k+1)}{2} \)
• \( \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \)
• \( \sum_{n=1}^{k} n^3 = \left( \frac{k(k+1)}{2} \right)^2 \)

Calculation:

Given,

\( D_n = \begin{bmatrix} n & 20 & 30 \ n^2 & 40 & 50 \ n^3 & 60 & 70 \ \end{bmatrix} \)

⇒ Extract common factors 20 and 10:

\( D_n = (20)(10) \begin{bmatrix} n & 1 & 3 \ n^2 & 2 & 5 \ n^3 & 3 & 7 \ \end{bmatrix} \)

⇒ Expand determinant using first column:

\( D_n = 200(-n + 2n^2 - n^3) \)

Now,

\( \sum_{n=1}^{4} D_n = 200 \sum_{n=1}^{4} (-n + 2n^2 - n^3) \)

⇒ \( = 200 \left[ -\frac{4(5)}{2} + 2 \cdot \frac{4(5)(9)}{6} - \left( \frac{4(5)}{2} \right)^2 \right] \)

⇒ \( = 200 (-10 + 60 - 100) \)

∴ The value is -10000.

Concept:

Matrix Multiplication and Properties:

• Matrix multiplication involves the dot product of rows and columns.
• A null matrix is a matrix in which all elements are zero.
• An identity matrix is a square matrix with 1's on the diagonal and 0's elsewhere.
• For matrices P and Q, PQ = QP does not generally hold unless P and Q commute.

Matrix Definitions:

• Null Matrix: A matrix where all elements are zero.
• Identity Matrix: A square matrix with 1's on the main diagonal and 0's elsewhere.

Calculation:

\(\rm P=\begin{bmatrix}0&c&-b\\ -c&0&a\\ b&-a&0\end{bmatrix}\ and\ \rm Q=\begin{bmatrix}a^2&ab&ac\\ ab&b^2&bc\\ ac&bc&c^2\end{bmatrix}\)

⇒ PQ = \(=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}\ \)

⇒QP = \(=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}\ \)

Then PQ = QP

∴ Option (c) is correct

Explanation:

The determinant of any skew-symmetric matrix with odd order is always zero

∴ Option (b) is correct.