Sequences & SeriesmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Sequences & Series: Let Roots Roots Equals (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question

Let

\alpha, \beta

be the roots of

x^2 - x + p = 0

and

\gamma, \delta

be the roots of

x^2 - 4x + q = 0

, where

p, q \in \mathbb{Z}

. If

\alpha, \beta, \gamma, \delta

are in G.P., then

|p + q|

equals:

16

32

34

correct

38

Solution

Step 1: Set up the G.P. Let

\alpha = a

\beta = ar

\gamma = ar^2

\delta = ar^3

for some common ratio

r \neq 0

. Step 2: Apply Vieta's formulas to both equations From

x^2 - x + p = 0

\alpha + \beta = a(1+r) = 1 \quad \cdots (1)

\alpha\beta = a^2r = p \quad \cdots (2)

From

x^2 - 4x + q = 0

\gamma + \delta = ar^2(1+r) = 4 \quad \cdots (3)

\gamma\delta = a^2r^5 = q \quad \cdots (4)

Step 3: Determine the common ratio Dividing equation (3) by equation (1):

\frac{ar^2(1+r)}{a(1+r)} = r^2 = \frac{4}{1} = 4 \implies r = \pm 2

Step 4: Test

r = 2

From (1):

a(1+2) = 1 \Rightarrow a = \frac{1}{3}

. Then

p = a^2r = \frac{1}{9}\cdot2 = \frac{2}{9} \notin \mathbb{Z}

. This case is rejected. Step 5: Test

r = -2

From (1):

a(1-2) = 1 \Rightarrow a = -1

. Verification via (3):

(-1)(4)(1-2) = (-1)(4)(-1) = 4

\checkmark

Step 6: Compute

p

q

, and the required value

p = a^2r = (-1)^2(-2) = -2, \qquad q = a^2r^5 = (1)(-2)^5 = -32

|p + q| = |-2 + (-32)| = 34

Answer: (3)

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