Question

Given A = {1, 2 40}. Three numbers are randomly selected from set A. Then, the probability that the terms form an increasing G.P is

Answer 1

7/2470

Answer 2

Solution:

We want to choose three numbers (in increasing order) from the set
  A = {1, 2, 3, …, 40}
so that they form a geometric progression (GP). Any increasing GP can be written as
  a, a·r, a·r²   (with r > 1)
Since all three terms must be in A, we require

  a·r² ≤ 40.

A systematic way is to “parameterize” r as a reduced rational number. Write
  r = p/q  (with integers p > q ≥ 1 and (p, q) = 1).
Then the GP is
  a, a·(p/q), a·(p/q)².
For the middle term to be an integer, a must be divisible by q; in fact, to make a·(p/q)² an integer we write
  a = q²·k,  k ∈ ℕ.

The three terms become
  q²·k,  p·q·k,  p²·k.

To have them in A we need
  p²·k ≤ 40.  (1)

For each eligible pair (q, p) (with p > q, (p, q) = 1) the number of valid k is
  k = 1, 2, …, ⎣40/p²⎦.

Now, summing over all possible (q, p):

1. For q = 1 (i.e. r = p):
     p² ≤ 40 ⟹ p = 2, 3, 4, 5, 6.
     • p = 2: ⎣40/4⎦ = 10
     • p = 3: ⎣40/9⎦ = 4
     • p = 4: ⎣40/16⎦ = 2
     • p = 5: ⎣40/25⎦ = 1
     • p = 6: ⎣40/36⎦ = 1
     Subtotal = 10 + 4 + 2 + 1 + 1 = 18.

2. For q = 2 (so a = 4k): p must be >2 and coprime to 2. Also p² ≤ 40 when multiplied by k = 1 is not automatic; rather the restriction is on p²·k. We count for k = 1 up to ⎣40/p²⎦.   • p = 3: ⎣40/9⎦ = 4
     • p = 5: ⎣40/25⎦ = 1
     Subtotal = 4 + 1 = 5.

3. For q = 3 (a = 9k): p > 3 and (p, 3) = 1.   • p = 4: ⎣40/16⎦ = 2
     • p = 5: ⎣40/25⎦ = 1
     Subtotal = 2 + 1 = 3.

4. For q = 4 (a = 16k): p > 4 and (p,4)=1.   • p = 5: ⎣40/25⎦ = 1
     Subtotal = 1.

5. For q = 5 (a = 25k): p > 5 and (p,5)=1.   • p = 6: ⎣40/36⎦ = 1
     Subtotal = 1.

For q ≥ 6, the smallest a = q² exceeds 40, so no further terms occur.

Total favorable GP triplets = 18 + 5 + 3 + 1 + 1 = 28.

The total number of ways to choose 3 numbers from A is
  C(40, 3) = (40·39·38)/6 = 9880.

Thus, the probability that three randomly chosen numbers form an increasing GP is
  Probability = 28/9880 = (28 ÷ 4)/(9880 ÷ 4) = 7/2470.

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Minimal Explanation:

1. Express any increasing GP as: a, a·(p/q), a·(p/q)² with a = q²·k (to ensure integrality).
2. For each (q,p) with p > q and (p,q)=1, k can take values from 1 to ⎣40/p²⎦.
3. Summing over all possibilities gives 28 favorable triplets.
4. Total number of triplets = C(40,3) = 9880.
5. Thus, probability = 28/9880 = 7/2470.

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Answer:
The required probability is  7/2470.