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

$\frac{7}{2470}$

Answer 2

Solution:

1. Total selections:
   Total ways to choose 3 numbers from set A = {1, 2, …, 40} is

   $$\binom{40}{3} = \frac{40 \times 39 \times 38}{6} = 9880.$$

2. Count of increasing geometric progressions:
   An increasing geometric progression has numbers $a, ar, ar^2$ with $r>1$ and all terms in A. We break into two cases:

   Case 1 (Integer $r$):
   For an integer $r \ge 2$, the condition $ar^2 \le 40$ implies

   • For $r=2$: $a \le \lfloor 40/4 \rfloor = 10$ → 10 sequences
   • For $r=3$: $a \le \lfloor 40/9 \rfloor = 4$ → 4 sequences
   • For $r=4$: $a \le \lfloor 40/16 \rfloor = 2$ → 2 sequences
   • For $r=5$: $a \le \lfloor 40/25 \rfloor = 1$ → 1 sequence
   • For $r=6$: $a \le \lfloor 40/36 \rfloor = 1$ → 1 sequence
     Total (integer $r$) = $10+4+2+1+1 = 18$.

   Case 2 (Rational non‐integer $r = \frac{p}{q}$ with $\gcd(p,q)=1$ and $p>q$, $q\ge 2$):
   Express the terms as:

   $$a = kq^2,\quad b = kpq,\quad c = kp^2,$$

   with $kp^2 \le 40$.
   Now, ensure $p^2 \le 40$ for $k=1$. Listing the possible pairs:

   • For $(p,q) = (3,2)$: $k \le \lfloor 40/9 \rfloor = 4$ → 4 sequences.
   • For $(p,q) = (4,3)$: $k \le \lfloor 40/16 \rfloor = 2$ → 2 sequences.
   • For $(p,q) = (5,2)$: $k \le \lfloor 40/25 \rfloor = 1$ → 1 sequence.
   • For $(p,q) = (5,3)$: $k \le \lfloor 40/25 \rfloor = 1$ → 1 sequence.
   • For $(p,q) = (5,4)$: $k \le \lfloor 40/25 \rfloor = 1$ → 1 sequence.
   • For $(p,q) = (6,5)$: $k \le \lfloor 40/36 \rfloor = 1$ → 1 sequence.
     Total (rational non‐integer) = $4+2+1+1+1+1 = 10$.

3. Total favorable outcomes: $18 + 10 = 28$.

4. Probability:

   $$\text{Probability} = \frac{28}{9880} = \frac{7}{2470}.$$

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Explanation (minimal):
Total ways = 9880. Count GP sequences by splitting into cases based on whether the common ratio $r$ is an integer or a non‐integer rational. Integer $r$ contributed 18 sequences and rational non‐integer $r$ contributed 10 sequences, totaling 28. Hence, probability = $\frac{28}{9880} = \frac{7}{2470}$.