Let α and β be the roots of x2+3x−16=0, and γ and δ be the roots of x2+3x−1=0. If Pn=αn+βn and Qn=γn+δn, then 2P23P25+3P24+Q24Q25−Q23 is equal to
(A)3
(B)4
(C)5✓ Correct
(D)7
Answer: (C)
Solution
From x2+3x−16=0 the sequence satisfies Pn+3Pn−1−16Pn−2=0, so P25+3P24=16P23 and 2P23P25+3P24=2P2316P23=8. From x2+3x−1=0, γ2−1=−3γ and δ2−1=−3δ, so Q25−Q23=γ23(γ2−1)+δ23(δ2−1)=−3(γ24+δ24)=−3Q24, giving Q24Q25−Q23=−3. The required value is 8+(−3)=5.
JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Single correct option