If a,b,c,d are in continued proportion then (a-d)/(b-c) >=x. What is the value of x.

a)2
b)3
c)0
d)1

• let a,b,c,d are a,ar,ar^2,ar^3 respectively such that a,b,c,d are continued proportion then
  
  (a-d)/(b-c)= (a-ar^3)/(ar-ar^2)=(1-r^3)/(r-r^2)= (1+r+r^2)/r = (1/r +r +1)
  
  since (r+ 1/r) is always greater or equal to 2
  so, (a-d)/(b-c)= (1/r +r +1) is always greater or equal to 3
  
  so, x=3
  
  b)3
• 10 years agoHelpfull: Yes(13) No(6)
• What if more than one choice was greater than 4
• 8 years agoHelpfull: Yes(0) No(0)
• If a,b, c,d are in continued proportion, a/b = b/c = c/d i.e a,b,c and d are in GP.
  
  We can say that:
  
  a = p, b=pr, c=pr^2 and d=pr^3 (p is the first term and r is the common ration of a GP series )
  
  (a-d)/(b-c) = p(1-r^3)/pr(1-r) => (1+r+r^2)/r => (1+r + 1/r)
  
  Minimum( 1+r + 1/r ) = 1 + Minimum( r + 1/r ) = 3. Hence value of x is 3.
• 6 years agoHelpfull: Yes(0) No(0)