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Calculation:

Given: $p^x = q^y = r^z = k$ and $x, y, z$ are in GP.

Taking logs: $\ln(p^x) = \ln(q^y) = \ln(r^z) = \ln k \Rightarrow x\ln p = y\ln q = z\ln r = A$

Thus: $\ln p = \frac{A}{x},\quad \ln q = \frac{A}{y},\quad \ln r = \frac{A}{z}$

Let $x, y, z$ be GP: $x=a,\ y=ar,\ z=ar^2$

Then: $\ln p = \frac{A}{a},\quad \ln q = \frac{A}{ar},\quad \ln r = \frac{A}{ar^2}$

So $\ln p,, \ln q,, \ln r$ are in GP (reciprocals of a GP form a GP).

Hence Statement II is true.

For Statement I (whether $p, q, r$ are in AP):

from $\ln p = \frac{A}{x},\ \ln q = \frac{A}{y},\ \ln r = \frac{A}{z}$, there is no general condition implying $q = \frac{p+r}{2}$;

logarithms of reciprocals of a GP do not enforce an AP on the original bases. Hence Statement I is false.

Hence the correct answer is II only.

Calculation:

Statement I:

$\text{L.H.S.} = A - B = A \cap B^{c}$

$\text{R.H.S.} = B^{c} - A^{c} = B^{c} \cap (A^{c})^{c} = B^{c} \cap A$

$\therefore ; \text{L.H.S.} = \text{R.H.S.}$

So, Statement I is correct.

Statement II:

$\text{L.H.S.} = A - B^{c} = A \cap (B^{c})^{c} = A \cap B$

$\text{R.H.S.} = A^{c} - B = A^{c} \cap B^{c} = U - (A \cup B)$

$\therefore ; \text{L.H.S.} \ne \text{R.H.S.}$

So, Statement II is incorrect.

Hence, the correct answer is Option 1.

Calculation:

Given: $y=x!,; z=(2x)!$ and $\dfrac{z}{y}=120$

⇒ $\dfrac{(2x)!}{x!}=120$

Check small integer values of $x$:

For $x=3$, $\dfrac{(2\cdot 3)!}{3!}=\dfrac{6!}{3!}=\dfrac{720}{6}=120$

So, $x=3$.

Now, $(3x)!=(3\cdot 3)!=9!=362880$

Hence the correct answer is 362880.

Calculation:

The number of trailing zeros in $n!$ depends on the number of factors of 5 in its prime factorization.

Number of trailing zeros $= \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \cdots$

We are given that the number of trailing zeros is exactly 2.

Checking possible values of n :

For $n = 10$:

⇒ $\left\lfloor \frac{10}{5} \right\rfloor = 2,; \left\lfloor \frac{10}{25} \right\rfloor = 0 \Rightarrow 2$

For $n = 11, 12, 13, 14$:

⇒ $\left\lfloor \frac{n}{5} \right\rfloor = 2,; \left\lfloor \frac{n}{25} \right\rfloor = 0 \Rightarrow 2$

For $n = 15$:

$\left\lfloor \frac{15}{5} \right\rfloor = 3 \Rightarrow$ 3 zeros (exceeds)

Hence, the valid values of $n$ are:

$n = 10, 11, 12, 13, 14$

Therefore, the number of possible values of n  is 5.

Calculation:

Given that:

$(10+\log_{10}x),\ (10+\log_{10}y),\ (10+\log_{10}z) \text{ are in A.P.}$

Using the condition of A.P.:

$2(10+\log_{10}y)=(10+\log_{10}x)+(10+\log_{10}z)$

Simplifying:

$20+2\log_{10}y=20+\log_{10}x+\log_{10}z$

$2\log_{10}y=\log_{10}x+\log_{10}z$

This implies:

$\log_{10}y=\frac{\log_{10}x+\log_{10}z}{2}$

Hence, the A.M. of $\log_{10}x$ and $\log_{10}z$ is $\log_{10}y$.

So, statement II is correct.

Now,

$2\log_{10}y=\log_{10}x+\log_{10}z \Rightarrow \log_{10}y^2=\log_{10}(xz)$

$\Rightarrow y^2=xz$

Therefore, the G.M. of $x$ and $z$ is $y$, not $y^2$.

So, statement I is incorrect.

Hence, the correct answer is Option 2.

Calculation:

The sum of the first $n$ odd natural numbers is given by:

$1 + 3 + 5 + \cdots + (2n-1) = n^2$

Given that the sum is:

$n^2 = 12345678987654321$

Taking square root on both sides:

$n = \sqrt{12345678987654321}$

Now,

$111111111^2 = 12345678987654321$

$\Rightarrow n = 111111111$

Hence, the correct answer is 111111111.

Calculation:

Given two Arithmetic Progressions (APs):

$\text{AP}_1: 19, 21, 23,$...... up to 110 terms

$a_1 = 19, d_1 = 2, T_n = 19 + 2(n-1) = 2n + 17$ (up to n = 110,where } T₁₁₀ = 237

$\text{AP}_2: 19, 22, 25,$...... up to 75 terms

$a_2 = 19, d_2 = 3, S_m = 19 + 3(m-1) = 3m + 16$ up to m = 75 , where  S₇₅ = 241

To find the common terms

$2n + 17 = 3m + 16 \Rightarrow 2n - 3m = -1 \Rightarrow 2n \equiv 2 \pmod 3 \Rightarrow n \equiv 1 \pmod 3$

$n = 3k + 1$, Now, we calculate the common terms

$T_{3k+1} = 2(3k+1) + 17 = 6k + 19$

Now, to ensure the terms are within the limits, we use

$6k + 19 \le 237 \Rightarrow 6k \le 218 \Rightarrow k \le 36$

Values of k  from 0 to 36 give 37 common terms

Hence, the correct answer is 37.

Calculation:

We are given that: $\alpha = \frac{-1 + \sqrt{3}}{2}$, which is the cube root of unity.

Using the properties of cube roots of unity:

⇒ $\alpha^3 = 1$

⇒ $1 + \alpha + \alpha^2 = 0$

⇒ $\alpha^2 = \alpha - 1$

Now, the given expression is:

⇒ $(1 + \alpha^{19} - \alpha^{35})^{100} - (1 - 3\alpha^{25} + \alpha^{38})^{50}$

Simplifying $\alpha^{19}$ and $\alpha^{35}$

⇒ $\alpha^{19} = \alpha$ since  $19 \mod 3 = 1$

⇒ $\alpha^{35} = \alpha^2$ Since $35 \mod 3 = 2$

Thus, $1 + \alpha^{19} - \alpha^{35} = 1 + \alpha - \alpha^2$

Using $\alpha^2 = \alpha - 1$, we get:

⇒ $1 + \alpha - (\alpha - 1) = 1 + \alpha - \alpha + 1 = 2$

So, $(1 + \alpha^{19} - \alpha^{35})^{100} = 2^{100}$

Simplifying $\alpha^{25}$ and $\alpha^{38}$

⇒ $\alpha^{25} = \alpha$ since $25 \mod 3 = 1$

⇒ $\alpha^{38} = \alpha^2$ Since $38 \mod 3 = 2$

Thus, $1 - 3\alpha^{25} + \alpha^{38} = 1 - 3\alpha + \alpha^2$

Using $\alpha^2 = \alpha - 1$, we get:

⇒ $1 - 3\alpha + (\alpha - 1) = 1 - 3\alpha + \alpha - 1 = -2\alpha$

So, $(1 - 3\alpha^{25} + \alpha^{38})^{50} = (-2\alpha)^{50} = 2^{50} \alpha^{50}$

Since $\alpha^3 = 1$, we have $\alpha^{50} = \alpha^2$.

Thus, $(1 - 3\alpha^{25} + \alpha^{38})^{50} = 2^{50} \alpha^2$

Subtracting the two terms

⇒ $2^{100} - 2^{50} \alpha^2$

Since $\alpha^2 = \alpha - 1$, we get:

⇒ $2^{100} - 2^{50} (\alpha - 1) = 2^{100} - 2^{50} \alpha + 2^{50}$

The value of the expression is $0$

Hence, the correct answer is Option 3.

Calculation:

We are asked to find the remainder when $5^{99}$ is divided by 13.

Let’s find the pattern of powers of 5 modulo 13:

$5^1 = 5 \mod 13 = 5$

$5^2 = 25 \mod 13 = 12$

$5^3 = 125 \mod 13 = 8$

$5^4 = 625 \mod 13 = 1$

So the powers repeat every 4 steps: $5^4 \equiv 1 \mod 13$

Now reduce the exponent modulo 4:

$99 \mod 4 = 3 \Rightarrow 5^{99} \equiv 5^3 \mod 13$

$5^3 = 125 \mod 13 = 8$

Hence, the remainder when $5^{99}$ is divided by 13 is 8.

Calculation:

We are asked to find the value of the determinant of the inverse of the matrix:

$\begin{bmatrix} -4 & -5 \ 2 & 2 \end{bmatrix}$

Compute the determinant of the original matrix:

⇒ $\text{det}(A) = (-4)(2) - (-5)(2) = -8 + 10 = 2$

Use the property of inverse determinant:

⇒ $\text{det}(A^{-1}) = \frac{1}{\text{det}(A)} = \frac{1}{2}$

Hence, the value of the determinant of the inverse is 1/2