numbers in G.P can be taken as a , ar ,a(r^2),where a is the 1st term and r is the common ratio
there are 100 numbers
CASE 1:
let us start by taking the r =2 ,therefore r^2 = 4
possible numbers from which choice of a can be made= 100/4 = 25
CASE 2:
let the next value of r =3 ,therefore r^2 = 9
possible numbers from which choice of a can be made=100/9 =11
{ we take only the maximum number divisible i.e the quotients}
CASE 3:
let the next value of r =4 ,therefore r^2 = 16
possible numbers from which choice of a can be made=100/16 =6
CASE 4:
let the next value of r =5 ,therefore r^2 = 25
possible numbers from which choice of a can be made=100/25 =4
CASE 5:
let the next value of r =6 AND r=7,therefore r^2 = 36 and 49 respectively
possible numbers from which choice of a can be made=100/36 +100/49= 4
{ takin only the quotient of the division}
CASE 6 :
when r=8, 9,10 the maximum numbers divisible is 1 each .since r^2=(64,81,100)
so total number is 1+1+1=3
WE DO NOT TAKE ANY FURTHER VALUES OF r SINCE IT WOULD EXIT THE LIMIT 0F 100
n(E) = event that the numbers in GP is the sum of the abve cases
= 25 + 11 +6 +4 +4+ 3 =53
n(S)= sample space of choosing 3 numbers is 100C3
probability = n(E)/n(S)
= 53/ 100C3
= 53/161700
Moderation Team
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