JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Probability: Three distinct numbers are selected randomly from the set ${1, 2, 3, \dots, 40}$ . If the probability, that the selected numbers are in an increasing G.P. is $\frac{m}{n}$ , $g cd (m, n) = 1$ , then $m + n$ is equal to _____.

Q: How do I know when to stop checking values for 'p'?

Great question. Look at the inequality $kp^2 \le 40$. Since $k$ must be at least $1$, the maximum possible value for $p^2$ is $40$. This means $p^2 \le 40$, so $p$ can only be $2, 3, 4, 5, 6$. Once $p=7$, $p^2=49$, which exceeds $40$. The math itself tells you exactly when to stop!

Q: Why is the condition $a = kq^2$ necessary? Can't I just pick any 'a'?

If you pick an arbitrary 'a', the term $ar^2$ might not be an integer. For example, if $a=1$ and $r=\frac{3}{2}$, the terms are $1, 1.5, 2.25$. None of these are integers! By setting $a = kq^2$, you guarantee that the denominator $q^2$ in the third term $a(\frac{p^2}{q^2})$ is cancelled out, ensuring all terms are integers. It is a structural requirement, not a choice.

Enter Numerical Value:

Visualized Solution

Total Sample Space $n (S)$

Total numbers in the set $S = {1, 2, 3, \dots, 40}$ .
Number of ways to select 3 distinct numbers is given by $n (S) =^{40} C_{3}$ .
$n (S) = \frac{40 \times 39 \times 38}{3 \times 2 \times 1} = 9880$ .

Defining the G.P. Condition

Let the selected numbers be $a, a r, a r^{2}$ in increasing order.
Constraints: $1 \leq a < a r < a r^{2} \leq 40$ .
Since the numbers are distinct and increasing, $r > 1$ .

The Rational Ratio Constraint

For $a, a r, a r^{2} \in Z$ , $r$ must be rational. Let $r = \frac{p}{q}$ where $g cd (p, q) = 1$ and $p > q \geq 1$ .
Terms: $a, a \frac{p}{q}, a \frac{p ^{2}}{q ^{2}}$ .
To ensure all terms are integers, $a$ must be a multiple of $q^{2}$ . Let $a = k q^{2}$ .
The G.P. becomes: $k q^{2}, k pq, k p^{2}$ .
Condition: $k p^{2} \leq 40$ .

Case 1: Common Ratio $r = 2$

Case 1: $p = 2, q = 1$ (Integer ratio $r = 2$ )
Condition: $k (2^{2}) \leq 40 \Rightarrow 4 k \leq 40 \Rightarrow k \in {1, 2, \dots, 10}$ .
Number of cases = 10.

Case 2: Common Ratio $r = 3$ and $r = \frac{3}{2}$

Case 2: $p = 3$ . $p^{2} = 9$ . Condition: $9 k \leq 40 \Rightarrow k \in {1, 2, 3, 4}$ .
If $q = 1$ ( $r = 3$ ): 4 cases.
If $q = 2$ ( $r = \frac{3}{2}$ ): 4 cases.
Total for $p = 3$ is $4 + 4 = 8$ cases.

Case 3: Common Ratio $r = 4$ and $r = \frac{4}{3}$

Case 3: $p = 4$ . $p^{2} = 16$ . Condition: $16 k \leq 40 \Rightarrow k \in {1, 2}$ .
Coprime $q < 4$ : $q \in {1, 3}$ .
If $q = 1$ ( $r = 4$ ): 2 cases.
If $q = 3$ ( $r = \frac{4}{3}$ ): 2 cases.
Total for $p = 4$ is $2 + 2 = 4$ cases.

Case 4: Common Ratio $r = 5$

Case 4: $p = 5$ . $p^{2} = 25$ . Condition: $25 k \leq 40 \Rightarrow k = 1$ .
Coprime $q < 5$ : $q \in {1, 2, 3, 4}$ .
Number of cases = 4 (one for each $q$ ).

Case 5: Common Ratio $r = 6$

Case 5: $p = 6$ . $p^{2} = 36$ . Condition: $36 k \leq 40 \Rightarrow k = 1$ .
Coprime $q < 6$ : $q \in {1, 5}$ .
Number of cases = 2 (one for each $q$ ).
Note: $p = 7 \Rightarrow p^{2} = 49 > 40$ , no cases possible.

Summing Favorable Outcomes $n (E)$

Total favorable cases $n (E) = 10 + 8 + 4 + 4 + 2$ .
$n (E) = 28$ .

Final Probability Calculation

Probability $P = \frac{n ( E )}{n ( S )} = \frac{28}{9880}$ .
Simplifying the fraction: $P = \frac{7}{2470}$ .
Given $P = \frac{m}{n}$ and $g cd (m, n) = 1$ .
Therefore, $m = 7$ and $n = 2470$ .

The Final Result

We need to find $m + n$ .
$m + n = 7 + 2470 = 2477$ .
Key Takeaway: For a G.P. in a set of integers, the common ratio $r$ can be a rational number $\frac{p}{q}$ . Always check for both integer and fractional ratios within the given bounds.

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The Sigma Insight: Classical Definition of Probability

Solution Diagram

Analyzing the Sample Space

The total number of ways to select three distinct numbers from the set

{1, 2, 3, \dots, 40}

is given by the combination formula

(3 40)

Calculating this value:

(3 40) = \frac{40 \times 39 \times 38}{3 \times 2 \times 1} = 9880

This value,

9880

, serves as the denominator for our probability calculation.

The Geometric Progression Condition

Let the three numbers in increasing Geometric Progression (G.P.) be

a, a r, a r^{2}

. For these to be integers, the common ratio

r

must be a rational number, which we can express as

r = \frac{p}{q}

in its simplest form (where

p > q

and

g cd (p, q) = 1

Substituting

r = \frac{p}{q}

into our terms, we get:

a, a (\frac{p}{q}), a (\frac{p ^{2}}{q ^{2}})

For all three terms to be integers,

a

must be a multiple of

q^{2}

. Let

a = k q^{2}

for some positive integer

k

. The terms then become

k q^{2}, k pq, k p^{2}

The Master Inequality

The constraint that all terms must be within the set

{1, 2, \dots, 40}

implies that the largest term must satisfy:

k p^{2} \leq 40

We now iterate through possible values of

p

and

q

(where

p > q

and

g cd (p, q) = 1

If $p = 2$ : $4 k \leq 40 \Rightarrow k \leq 10$ . Since $q = 1$ , we have 10 cases. If $p = 3$ :

9 k \leq 40 \Rightarrow k \leq 4

. Since

q

can be

1

2

, we have

4 + 4 =

8 cases. If $p = 4$ : $16 k \leq 40 \Rightarrow k \leq 2$ . Since $q$ can be $1$ or $3$ (note $\gcd(4, 2) eq 1$), we have $2 + 2 =$ 4 cases. If $p = 5$ :

25 k \leq 40 \Rightarrow k = 1

. Since

q

can be

1, 2, 3, 4

, we have 4 cases. If $p = 6$ : $36 k \leq 40 \Rightarrow k = 1$ . Since $q$ can be $1$ or $5$ (note $\gcd(6, 2) eq 1, \gcd(6, 3) eq 1, \gcd(6, 4) eq 1$), we have 2 cases*.

Final Calculation

Summing the valid cases identified above:

10 + 8 + 4 + 4 + 2 = 28

The probability is therefore:

P = \frac{28}{9880} = \frac{7}{2470}

Given

m = 7

and

n = 2470

, the final result is

m + n = 7 + 2470 = 2477

Common Traps & FAQs

Wait, why can't we just assume the common ratio is an integer? Isn't that simpler?

That is the classic trap! If you only consider integer ratios like

r = 2

r = 3

, you will miss sequences like

4, 6, 9

(where

r = 1.5

\frac{3}{2}

). In JEE Advanced, they test your ability to generalize. Always remember that a G.P. is defined by the ratio of consecutive terms, and that ratio can absolutely be a rational fraction.

How do I know when to stop checking values for 'p'?

Great question. Look at the inequality

k p^{2} \leq 40

. Since

k

must be at least

1

, the maximum possible value for

p^{2}

40

. This means

p^{2} \leq 40

, so

p

can only be

2, 3, 4, 5, 6

. Once

p = 7

p^{2} = 49

, which exceeds

40

. The math itself tells you exactly when to stop!

Why is the condition

a = k q^{2}

necessary? Can't I just pick any 'a'?

If you pick an arbitrary 'a', the term

a r^{2}

might not be an integer. For example, if

a = 1

and

r = \frac{3}{2}

, the terms are

1, 1.5, 2.25

. None of these are integers! By setting

a = k q^{2}

, you guarantee that the denominator

q^{2}

in the third term

a (\frac{p ^{2}}{q ^{2}})

is cancelled out, ensuring all terms are integers. It is a structural requirement, not a choice.

Animated Solution for Mathematics - Probability: Three distinct numbers are selected randomly from the set ${1, 2, 3, \dots, 40}$ . If the probability, that the selected numbers are in an increasing G.P. is $\frac{m}{n}$ , $g cd (m, n) = 1$ , then $m + n$ is equal to _____.

Enter Numerical Value:

Visualized Solution

Total Sample Space $n (S)$

Defining the G.P. Condition

The Rational Ratio Constraint

Case 1: Common Ratio $r = 2$

Case 2: Common Ratio $r = 3$ and $r = \frac{3}{2}$

Case 3: Common Ratio $r = 4$ and $r = \frac{4}{3}$

Case 4: Common Ratio $r = 5$

Case 5: Common Ratio $r = 6$

Summing Favorable Outcomes $n (E)$

Final Probability Calculation

The Final Result

The Sigma Insight: Classical Definition of Probability

Analyzing the Sample Space

The Geometric Progression Condition

The Master Inequality

Final Calculation

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Comprehension Passage

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Enter Numerical Value:

Visualized Solution

Total Sample Space .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(S)

Defining the G.P. Condition

The Rational Ratio Constraint

Case 1: Common Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=2

Case 2: Common Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=3 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=23​

Case 3: Common Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=4 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=34​

Case 4: Common Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=5

Case 5: Common Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=6

Summing Favorable Outcomes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(E)

Final Probability Calculation

The Final Result

The Sigma Insight: Classical Definition of Probability

Analyzing the Sample Space

The Geometric Progression Condition

The Master Inequality

Final Calculation

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Total Sample Space $n (S)$

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