Cubes c1,c2,c3.....and Spheres s1,s2,s3,....are defined in the following way.
-s1 has radius 4cm.
-For each n>0, Cn is inscribed in Sn and Sn+1 is inscribed in Cn(ie c1 is inscribed in s1,s2 is inscribed in c1,c2 is inscribed in s2 and so on)
Let Vn be the sum of the volumes of the first n cubes c1,c2,c3....Cn.Then as n->infinity Vn approaches.

options i)384(1+3root3)/13 2)64(1+3root3)/13 3) 128(1+3root3)/13 4) infinity , ie Vn is unbounded.

• c is the right answer.
• 12 years agoHelpfull: Yes(4) No(10)
• There is no option here
  Ans is 256(1 + 3sqrt3)/13
  
  In a sphere of radius R , length of cube inscribed = 2R / sqrt3
  => V1 = (2R / sqrt3)^3
  
  Radius of sphere inscribed in this cube = R / sqrt3
  cube inside S2 = 2R / 3
  => V2 = (2R / 3)^3 and so on
  
  Therefore V1 + V2 + . . . infinity
  = (2R)^3 [ (1/sqrt)^3 + (1/3)^3 + (1/3sqrt3)^3 + . . . Infinity ]
  
  Sum of infinite terms = first term / ( 1 - common ratio)
  = 1 / (3sqrt3 - 1) = (1 + 3sqrt3) / 13
  
  => V = (2R)^3 ( 1 + 3sqrt3) / (13 x 2)
  Here R = 4 cm
  => V = 256(1 + 3sqrt3) / 13
  
  
  
  
• 12 years agoHelpfull: Yes(1) No(0)
• ans is 256(1+3root3)/13
  for cube of side a inserted in a sphere radius r
  relation b/w a && r is a=2r/root3
  for sphere of radius x inserted in a cube of side b
  relation b/w b && x is x=b/2
  by using above relations sides of the cube c1,c2,c3,...... are in gp with initial value 2r/root3 and common ratio 1/root3
• 12 years agoHelpfull: Yes(0) No(0)