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 α,β\alpha, \beta be the roots of x2−x+p=0x^2 - x + p = 0 and γ,δ\gamma, \delta be the roots of x2−4x+q=0x^2 - 4x + q = 0, where p,q∈Zp, q \in \mathbb{Z}. If α,β,γ,δ\alpha, \beta, \gamma, \delta are in G.P., then ∣p+q∣|p + q| equals:
A1616
B3232
C3434correct
D3838
Solution
Step 1: Set up the G.P. Let α=a\alpha = a, β=ar\beta = ar, γ=ar2\gamma = ar^2, δ=ar3\delta = ar^3 for some common ratio r≠0r \neq 0. Step 2: Apply Vieta's formulas to both equations From x2−x+p=0x^2 - x + p = 0:
α+β=a(1+r)=1⋯(1)\alpha + \beta = a(1+r) = 1 \quad \cdots (1)
αβ=a2r=p⋯(2)\alpha\beta = a^2r = p \quad \cdots (2)
 From x2−4x+q=0x^2 - 4x + q = 0:
γ+δ=ar2(1+r)=4⋯(3)\gamma + \delta = ar^2(1+r) = 4 \quad \cdots (3)
γδ=a2r5=q⋯(4)\gamma\delta = a^2r^5 = q \quad \cdots (4)
 Step 3: Determine the common ratio Dividing equation (3) by equation (1):
ar2(1+r)a(1+r)=r2=41=4  ⟹  r=±2\frac{ar^2(1+r)}{a(1+r)} = r^2 = \frac{4}{1} = 4 \implies r = \pm 2
 Step 4: Test r=2r = 2 From (1): a(1+2)=1⇒a=13a(1+2) = 1 \Rightarrow a = \frac{1}{3}. Then p=a2r=19⋅2=29∉Zp = a^2r = \frac{1}{9}\cdot2 = \frac{2}{9} \notin \mathbb{Z}. This case is rejected. Step 5: Test r=−2r = -2 From (1): a(1−2)=1⇒a=−1a(1-2) = 1 \Rightarrow a = -1. Verification via (3): (−1)(4)(1−2)=(−1)(4)(−1)=4(-1)(4)(1-2) = (-1)(4)(-1) = 4. ✓\checkmark Step 6: Compute pp, qq, and the required value
p=a2r=(−1)2(−2)=−2,q=a2r5=(1)(−2)5=−32p = a^2r = (-1)^2(-2) = -2, \qquad q = a^2r^5 = (1)(-2)^5 = -32
∣p+q∣=∣−2+(−32)∣=34|p + q| = |-2 + (-32)| = 34
 Answer: (3)
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