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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
α + β = 1, αβ = p
γ + δ = 4, γ + δ = q
Since α, β, γ, δ are in GP, then let us assume them to be equal to α, αr, αr2, αr3 respectively.
∴ α + αr = 1 and αr2 + αr3= 4
α(r + 1) = 1 and αr2 (r + 1) = 4
Diving both, we get r2 = 4
r = ± 2
If r = 2
(Not possible as P and Q are integer)
∴r = – 2
α = – 1, β = 2, γ = – 4, δ = 8
P = – 2, Q = – 34
P + Q = – 34