Question

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

Answer 1

Let the three numbers in GP: a/r, a, ar---------(A)
Where '^' is power of . . . 

Sum of these numbers are: 
a/r +a +ar = 38 
a(1\r+1+r) = 38 ---- (1) 
Product of these numbers are: 
a^3 = 1728 
= (12)^3 
a = 12 

Putting the value of a in (1) you will get: 

12(1\r+1+r) = 38 

And factorising, we get 
r = 2/3 or r = 3/2

Sub. the r and a value in (A), we get 

8,12,18 or 18,12,8. When a = 12 

And smallest no. is 8.
Note:How do I realise its three terms
It's actually not easy
But if you know that 1728 is the cube of 12
So u actually can make out that the gp may have 3 terms
Hope it helps
:)

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.