Find the smallest number in a GP whose sum is 38 and product 1728

(a) 12 (b) 20 (c) 8 (d) none of these

• In GP :- a/r+a+ar....=38
  a(1+r+r^2)=38r......eq1
  Product a^3=1728
  a=12
  now put this value in eq1
  12(1+r+r^2)=38r
  r=2/3 and 3/2..so sallest number is 8,12,20
• 12 years agoHelpfull: Yes(31) No(11)
• Gp is 8,12,18.
  So smallest no. Is 8.
• 12 years agoHelpfull: Yes(13) No(11)
• let, GP: a/r+a+a+ar..=38
  take a/r as common
  a/r(1++r^2+...)=38
  a(1++r^2+...)=38r
  We know that product in GP= a^3
  a^3=1728
  a=12
  sub in the above equation
  12(1+r+r^2...)=38r
  solving r=3/2 and 2/3
  now 12/r=12/(3/2)=8
  Hence... Option C)8
• 4 years agoHelpfull: Yes(0) No(0)