Let a1a2,b1b2,c1c2 be the consecutive terms of an arithmetic progression. If a1x2+2b1x+c1=0anda2x2+2b2x+c2=0 have a common root, then a2,b2,c2 are in
A
A.P.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
G.P.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
H.P.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
A.G.P.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is BG.P. Let α is the common root a1α2+2b1α+c1=0 a2α2+2b2α+c2=0 Solving as a simultaneous equation α22(b1c2−b2c1)=α(c1a2−c2a1)=12(a1b2−a2b1)