Question

Let a,b,c,d and p be any non zero distinct real numbers such that (a² + b² + c²)p² – 2(ab + bc + cd)p + (b² + c² + d²) = 0. Then : 

(1) a,c,p are in G.P. 

(2) a,c,p are in A.P. 

(3) a,b,c,d are in G.P. 

(4) a,b,c,d are in A.P. 

Answer 1

(3) a,b,c,d are in G.P. 

(a² + b² + c²)p² + 2(ab + bc + cd)p + b² + c² + d² = 0 

(a²p² + 2abp + b²) + (b²p² + 2bcp + c²) + (c²p² + 2cdp + d²) = 0 

(ab + b)² + (bp + c)² + (cp + d)² = 0 

This is possible only when 

ap + b = 0 and bp + c = 0 and cp + d = 0 

p = -b/a = -c/b = -d/c

or b/a = c/b = d/c

∴ a,b,c,d are in G.P.