Given:
pᵗʰ term = q. (1)
qᵗʰ term = p. (2)

Let a₁ be first term and d be common difference.
Using (1):
a₁ + (p-1)d = q. (3)

Using (2):
a₁ + (q-1)d = p. (4)

Subtracting (4) from (3):
(p-q)d = q-p
d = -1

Substituting d = -1 in (3):
a₁ + (p-1)(-1) = q
a₁ – p + 1 = q
a₁ = p + q – 1

Now, (p+q)ᵗʰ term = a₁ + (p+q-1)(-1)
= (p+q-1) + (p+q-1)(-1)
= p+q-1 – p-q+1
= 0

Therefore, (p+q)ᵗʰ term is 0 is the correct answer.

Given first three terms a₁, a₂, a₃ of A.P. are:
a₁ = 3y – 1
a₂ = 3y + 5
a₃ = 5y + 1
In an A.P., a₂ – a₁ = a₃ – a₂

So:
(3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)

Simplifying LHS:
3y + 5 – 3y + 1 = 6

Simplifying RHS:
5y + 1 – 3y – 5 = 2y – 4

As LHS = RHS:
6 = 2y – 4

Adding 4 to both sides:
10 = 2y

Thus:
y = 2
Thus, 2 is the correct answer.

Given quadratic equation: x² – (a + 1)x + a = 0

For equal roots, discriminant should be zero:
b² – 4ac = 0

Here:
a = 1 (coefficient of x²)
b = -(a + 1)
c = a

Substituting in discriminant:
[-(a + 1)]² – 4(1)(a) = 0
(a + 1)² – 4a = 0
a² + 2a + 1 – 4a = 0
a² – 2a + 1 = 0
(a – 1)² = 0

Therefore:
a = 1

This can be verified by substituting a = 1 in original equation:
x² – 2x + 1 = 0
(x – 1)² = 0
x = 1 (repeated root)

Hence, 1 is the correct answer.

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