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Given:

The sum of the first k terms of a series S is given as S_k = 3k² + 5k.

Concept:

To find the nth term of a series, we use the formula:

a_n = S_n - S_n-1

If the nth term forms an arithmetic progression (AP), the common difference (d) is given by:

d = a_n+1 - a_n

Calculation:

We are given S_k = 3k² + 5k.

⇒ a_n = S_n - S_n-1

Also S_n = 3n² + 5n and S_n-1 = 3(n-1)² + 5(n-1)

⇒ a_n = [3n² + 5n] - [3(n-1)² + 5(n-1)]

⇒ a_n = [3n² + 5n] - [3(n² - 2n + 1) + 5n - 5]

⇒ a_n = [3n² + 5n] - [3n² - 6n + 3 + 5n - 5]

⇒ a_n = 3n² + 5n - 3n² + 6n - 3 - 5n + 5

⇒ a_n = 6n + 2

The series is an arithmetic progression (AP) if the difference between consecutive terms is constant.

⇒ Common difference d = a_n+1 - a_n

⇒ a_n+1 = 6(n+1) + 2 = 6n + 6 + 2 = 6n + 8

⇒ d = (6n + 8) - (6n + 2) = 6

∴ The terms of the series form an arithmetic progression with a common difference of 6.

Hence, the correct answer is Option B.

Concept:

Geometric Progression (GP) sums:

• First term = a.
• Common ratio = r (r ≠ 1).
• Sum of first n terms: S_n = a × (1 − rⁿ) / (1 − r).
• Key idea: use formula for n = 8 and n = 4.
• Algebra trick: substitute y = r⁴ to simplify.

 

Calculation:

Given relation between sums

⇒ S₈ = 5 × S₄

Using sum formula for r ≠ 1

⇒ a(1 − r⁸)/(1 − r)

⇒ = 5 × a(1 − r⁴)/(1 − r)

Cancel a and (1 − r)

⇒ 1 − r⁸ = 5(1 − r⁴)

Bring all terms to one side

⇒ r⁸ − 5r⁴ + 4 = 0

Put y = r⁴ (so y ≥ 0 for real r)

⇒ y² − 5y + 4 = 0

Factor the quadratic

⇒ (y − 1)(y − 4) = 0

So y = 1 or y = 4

Case 1: y = 1 ⇒ r⁴ = 1

⇒ real r are 1 or −1

⇒ but r ≠ 1 (given), so keep r = −1

Case 2: y = 4 ⇒ r⁴ = 4

⇒ real r are √2 and −√2

Collect real values (excluding r = 1)

⇒ r = −1, r = √2, r = −√2

∴ Number of possible real values of r = 3.

Given:

The quadratic equation is x² - kx + k = 0.

One root exceeds the other by 2√3.

⇒ α - β = 2√3.

Also, 

Sum of roots: α + β = k

Product of roots: α × β = k

Calculation:

We know the following identity 

$(\alpha + \beta )^2 = (\alpha - \beta)^2 + 4\alpha\beta$

⇒ k² = (2√3)² - 4k 

⇒ k² - 12 - 4k = 0

⇒ k² - 6k + 2k -12 = 0

⇒ k(k - 6) + 2 ( k - 6) = 0

⇒ (k - 6) (k + 2) = 0

⇒ k = 6 and k = -2

Thus, the possible values of k are 6 and -2

Hence, the correct answer is Option 2.

Calculation:

$x+\frac{5}{y}=4$

$xy + 5 = 4y .....(1)$

$y+\frac{5}{x}= -4$

$xy + 5 = -4x .... (2)$

Subtracting 1 and 2

$(xy+5)−(xy+5)=(4y)−(−4x)$

$0=4y+4x$

$x+y=0.$

Hence, the correct answer is Option 1.

Concept:

In an AP, the nth term is given by:

T_n = a + (n - 1)d

In a GP, terms satisfy the relation:

T_m² = T_n × T_p

Calculation:

Let the 5^th, 7^th, and 13^th terms of the AP be T₅, T₇, and T₁₃ respectively.

⇒ T₅ = a + 4d

⇒ T₇ = a + 6d

⇒ T₁₃ = a + 12d

Since the terms are in GP:

⇒ T₇² = T₅ × T₁₃

⇒ (a + 6d)² = (a + 4d) × (a + 12d)

Expanding both sides:

⇒ (a + 6d)² = a² + 12ad + 48d²

⇒ a² + 12ad + 36d² = a² + 16ad + 48d²

Cancel out common terms:

⇒ 36d² = 48d² + 4ad

⇒ -12d² - 4ad = 0

⇒ -4d( 3d + a) = 0

Since d ≠ 0

⇒ a = -3d

Conclusion:

∴ Ratio of the first term to the common difference is:

⇒ a/d = -3

Hence, the correct answer is Option 1.

Concept:

1. Arithmetic Progression (AP): The difference between consecutive terms is constant.

⇒ If p, a, q are in AP, then 2a = p + q.

2. Geometric Progression (GP): The ratio of consecutive terms is constant.

⇒ If p, b, q are in GP, then b² = pq.

3. Harmonic Progression (HP): Reciprocal of terms are in AP.

⇒ If p, c, q are in HP, then (1/p), (1/c), (1/q) are in AP.

Calculation:

Using AP condition: p, 1, q are in AP.

⇒ 2(1) = p + q

⇒ p + q = 2 ............(1)

Using GP condition: p, 2, q are in GP.

⇒ 2² = p × q

⇒ 4 = pq ............(2)

Now,

$\frac{2pq}{p + q} = \frac{8}{2} = 4$

and 

$\frac{p + q}{2pq} = \frac{1}{4}$

∴  (1/p), (1/4), (1/q) are in AP and 

 p, 4 and q are in H.P.

Thus, both statements are true.

Hence, the correct answer is Option 3.

Calculation:

Convert x, y, and z from binary to decimal:

⇒ x = (1111)₂ = $1\cdot2^{3}+1\cdot2^{2}+1\cdot2^{1}+1\cdot2^{0}=15$

⇒ y = (1001)₂ = $1\cdot2^{3}+0\cdot2^{2}+0\cdot2^{1}+1\cdot2^{0}=9$

⇒ z = (110)₂ = $1\cdot2^{2}+1\cdot2^{1}+0\cdot2^{0}=6$

Now evaluate  $x^{3}-y^{3}-z^{3}-3xyz$

$15^{3}-9^{3}-6^{3}-3!\times!15!\times!9!\times!6 = 3375-729-216-2430 = 0$

The result of the expression is 0. In binary, 0 is simply (0)₂.

Hence, the correct answer is Option 4.

Calculation:

Cofactors:

⇒ A = ei – hf

⇒ B = gf – di,

⇒ C = dh – ge

⇒ D = ch – bi and

⇒ G = bf – ec

Now, bB + cC – dD – gG 

= bgf – bdi + cdh – cge – cdh + bdi – gbf + gec = 0

Hence, the correct answer is Option 1.

Calculation:

Given,

Δ  =  $\begin{vmatrix} k(k+2) & 2k+1 & 1\ 2k+1 & k+2 & 1\ 3 & 3 & 1 \end{vmatrix}$

Simplify the determinant by the column operation  $C_1 \rightarrow C_1 - C_2$:

$\Delta = \begin{vmatrix} k^{2}-1 & 2k+1 & 1\ k-1 & k+2 & 1\ 0 & 3 & 1 \end{vmatrix}$

Expanding along the third row,

$\Delta = -3 \begin{vmatrix} k^{2}-1 & 1\ k-1 & 1 \end{vmatrix} ;+; \begin{vmatrix} k^{2}-1 & 2k+1\ k-1 & k+2 \end{vmatrix} = (k-1)^{3}.$

Thus  $\Delta = (k-1)^{3}$.

Sign analysis

• $k>0$: if $0<k<1$, Δ < 0; if $k>1$, Δ > 0  ⇒ Statement I is false.
• $k<0$:  $\Delta<0$  ⇒ Statement II is true.
• $k=0$:  $\Delta=(-1)^{3}=-1\neq0$  ⇒ Statement III is false.

∴ Only Statement II is correct  ⇒  exactly one statement is true.

Hence, the correct answer is Option 2.

Calculation:

Determinant Δ = $a(ei−fh)−b(di−fg)+c(dh−eg)$

Now, For our matrix, 

$a=2,b=3+i,c=−1,d=3−i,e=0,f=i,g=−1,h=−i,i=1$

calculate the subdeterminants

⇒ $ei−fh=(0)(1)−(i)(−i)=0−(−1)=1$

⇒ $di−fg=(3−i)(1)−(i)(−1)=3−i+i=3$

⇒ $dh−eg=(3−i)(−i)−(0)(−1)=−3i+i 2=−3i−1=−1−3i$

⇒ Δ = $2(1)−(3+i)(3)+(−1)(−1−3i)$

⇒ Δ = $2−9−3i+1+3i$

⇒ $Δ=−6+0i$

Since we are given that $Δ=A+iB$ comparing the real and imaginary parts, we find:

A  = -6 and B = 0

Thus A + B = -6 + 0 = - 6

Hence, the Correct answer is Option 2.