Question

find the smallest number in a gp whose sum is 38 and product 1728.

Answer 1

Let x,y,z be the numbers in geometric progression.

y^2=xz 
x+y+z=38 

xyz=1728 
xyz = xzy = y^2y = y^3 = 1728 

y = 12 
y^2=xz=144 
z=144/x 

x+y+z = x+12+144/x = 38 
x^2+12x+144=38x 
x^2-26x+144=0 
(x-18)(x-8)=0 
x=8,18 
If x =8, z = 38-8-12=18 
The numbers are 8,12, 18 

Their sum is 38
Their product is 1,728 

The smallest number is 8

Answer 2

Answer:

Let a/r, a, and ar be the three numbers in GP.

Sum, a/r + a + ar = 38 …(i)

Product, (a/r)a(ar) = 1728

a³= 1728

Taking cube root

a = 12

Substitute a in (i)

(12/r) + 12 + 12r = 38

(12/r) + 12r = 26

((1/r) + r) = 26/12

(r² + 1)/ r = 13/6

6r²-13r+6 = 0

Solving using the quadratic formula, we get

r = 2/3or 3/2

The numbers will be 18, 12, 8 or 18, 12, 8.

The smallest number is 8.