Question 289 of 519HardMCQ
Let a,b,c,d,p be any nonzero distinct real numbers such that (a2+b2+c2)p2−2(ab+bc+cd)p+(b2+c2+d2)=0. Then: Solutions · 1
Official solution
Rearranging: (ap−b)2+(bp−c)2+(cp−d)2=0, so each term is zero:
\[ap=b,\ bp=c,\ cp=d \implies \frac ba=\frac cb=\frac dc=p.\]
So a,b,c,d are in G.P. Answer the question to unlock