JEE Main 2025 April

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Animated Solution for Mathematics - Sequence and Series: Let $x_{1}, x_{2}, x_{3}, x_{4}$ be in a geometric progression. 2, 7, 9, 5 are subtracted respectively from $x_{1}, x_{2}, x_{3}, x_{4}$ then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4})$ is :

Q: Why can't we just use the sum of AP formula to solve this?

That is a great question! The sum of an AP formula, $S_n = \frac{n}{2}(2a + (n-1)d)$, requires knowing the common difference $d$ and the number of terms. Here, we don't know $d$ directly. The property $2b = a+c$ is much more powerful because it links the terms directly without needing to introduce $d$ as an extra variable.

Q: Is it possible for $r$ to be 1?

Let's test it! If $r=1$, then the GP terms are $a, a, a, a$. The AP terms would be $a-2, a-7, a-9, a-5$. For these to be in AP, the common difference must be constant. Here, $(a-7)-(a-2) = -5$, but $(a-9)-(a-7) = -2$. Since $-5 \neq -2$, $r$ cannot be 1. Our algebraic result $a(r-1)^2 = -3$ also confirms this, as $0$ cannot equal $-3$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the sequence. The core of this problem lies in the definition of an Arithmetic Progression (AP). Recall that for any three consecutive terms

a, b, c

in an AP, the property

2 b = a + c

must hold. Apply this to the first three and last three terms of your new sequence.

Hint 2

Hint 2: Look for algebraic patterns. When you set up the two equations from the AP property, you will notice a common quadratic expression involving

r

appearing in both. Specifically, the term

(r - 1)^{2}

will emerge. Do not expand it fully; keep it factored to make the next step easier.

Hint 3

Hint 3: The Execution. Start by writing the two equations:

2 (a r - 7) = (a - 2) + (a r^{2} - 9)

and

2 (a r^{2} - 9) = (a r - 7) + (a r^{3} - 5)

. Simplify these to

a (r - 1)^{2} = - 3

and

a r (r - 1)^{2} = - 6

. Dividing the second by the first will instantly reveal the value of

r

Visualized Solution

Define the G.P. Terms

Let the terms in G.P. be: $x_{1} = a$ , $x_{2} = a r$ , $x_{3} = a r^{2}$ , $x_{4} = a r^{3}$

Forming the A.P.

Subtracting $2, 7, 9, 5$ respectively:
New terms: $(a - 2), (a r - 7), (a r^{2} - 9), (a r^{3} - 5)$ are in A.P.

First A.P. Condition

For the first three terms in A.P.: $2 b = a + c$
$2 (a r - 7) = (a - 2) + (a r^{2} - 9)$

Simplify First Equation

Expand: $2 a r - 14 = a + a r^{2} - 11$
Rearrange: $a r^{2} - 2 a r + a = - 3$

Factorize First Equation

Factor out $a$ : $a (r^{2} - 2 r + 1) = - 3$
$\Rightarrow a (r - 1)^{2} = - 3$ ... (Equation 1)

Second A.P. Condition

For the last three terms in A.P.:
$2 (a r^{2} - 9) = (a r - 7) + (a r^{3} - 5)$

Simplify Second Equation

Expand: $2 a r^{2} - 18 = a r + a r^{3} - 12$
Rearrange: $a r^{3} - 2 a r^{2} + a r = - 6$

Factorize Second Equation

Factor out $a r$ : $a r (r^{2} - 2 r + 1) = - 6$
$\Rightarrow a r (r - 1)^{2} = - 6$ ... (Equation 2)

Solve for Common Ratio $r$

Divide Equation 2 by Equation 1:
$\frac{a r ( r - 1 ) ^{2}}{a ( r - 1 ) ^{2}} = \frac{- 6}{- 3}$
$\Rightarrow r = 2$

Calculate First Term $a$

Substitute $r = 2$ into Equation 1:
$a (2 - 1)^{2} = - 3$
$\Rightarrow a (1)^{2} = - 3 \Rightarrow a = - 3$

The Product Expression

We need $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4})$
Product $P = (a) (a r) (a r^{2}) (a r^{3}) = a^{4} r^{6}$

Substitute and Calculate

Substitute $a = - 3$ and $r = 2$ :
$P = (- 3)^{4} \cdot (2)^{6} = 81 \times 64 = 5184$

Final Division

Final Value $= \frac{1}{24} \times 5184$
Final Value $= 216$

00:00 / 00:00

The Sigma Insight: Geometric Progression (G.P.)

Solution Diagram

Analyzing the Setup

We are given four numbers

x_{1}, x_{2}, x_{3}, x_{4}

in a Geometric Progression (GP). We define them as:

x_{1} = a

x_{2} = a r

x_{3} = a r^{2}

, and

x_{4} = a r^{3}

The Shift to Arithmetic

When we subtract

2, 7, 9, 5

from these terms respectively, the resulting sequence forms an Arithmetic Progression (AP). The new terms are:

(a - 2), (a r - 7), (a r^{2} - 9), (a r^{3} - 5)

For any three consecutive terms in an AP, the middle term is the arithmetic mean of its neighbors. This is expressed as:

2 \times middle = first + third

The Algebraic Bridge

Applying this to the first three terms:

2 (a r - 7) = (a - 2) + (a r^{2} - 9)

Expanding and rearranging the terms:

2 a r - 14 = a + a r^{2} - 11

a r^{2} - 2 a r + a = - 3

Recognizing the perfect square on the left side:

a (r^{2} - 2 r + 1) = - 3

a (r - 1)^{2} = - 3 \dots (Equation 1)

Now, applying the same logic to the last three terms:

2 (a r^{2} - 9) = (a r - 7) + (a r^{3} - 5)

Expanding and rearranging:

2 a r^{2} - 18 = a r + a r^{3} - 12

a r^{3} - 2 a r^{2} + a r = - 6

Factoring out

a r

a r (r^{2} - 2 r + 1) = - 6

a r (r - 1)^{2} = - 6 \dots (Equation 2)

The Elegant Cancellation

We now have two equations:

a (r - 1)^{2} = - 3

a r (r - 1)^{2} = - 6

Dividing Equation 2 by Equation 1, the terms

a

and

(r - 1)^{2}

cancel out:

r = \frac{- 6}{- 3} = 2

Substituting

r = 2

back into Equation 1:

a (2 - 1)^{2} = - 3

a = - 3

The Final Victory

The problem asks for the value of

\frac{1}{24} (x_{1} x_{2} x_{3} x_{4})

. The product of the GP terms is:

P = (a) (a r) (a r^{2}) (a r^{3}) = a^{4} r^{6}

Substituting our values

a = - 3

and

r = 2

P = (- 3)^{4} \cdot (2)^{6} = 81 \times 64 = 5184

Finally, dividing by

24

\frac{5184}{24} = 216

The final result is 216.

Common Traps & FAQs

Why can't we just use the sum of AP formula to solve this?

That is a great question! The sum of an AP formula,

S_{n} = \frac{n}{2} (2 a + (n - 1) d)

, requires knowing the common difference

d

and the number of terms. Here, we don't know

d

directly. The property

2 b = a + c

is much more powerful because it links the terms directly without needing to introduce

d

as an extra variable.

Why did we divide the two equations instead of substituting?

You could substitute, but dividing is a 'pro' move in JEE problems! By dividing

a r (r - 1)^{2} = - 6

a (r - 1)^{2} = - 3

, the complex term

(r - 1)^{2}

and the variable

a

cancel out immediately. It is the most elegant path to finding

r

without getting bogged down in messy algebra.

Is it possible for

r

to be 1?

Let's test it! If

r = 1

, then the GP terms are

a, a, a, a

. The AP terms would be

a - 2, a - 7, a - 9, a - 5

. For these to be in AP, the common difference must be constant. Here,

(a - 7) - (a - 2) = - 5

, but

(a - 9) - (a - 7) = - 2

. Since

- 5 \neq = - 2

r

cannot be 1. Our algebraic result

a (r - 1)^{2} = - 3

also confirms this, as

0

cannot equal

- 3

Select Answer:

Visualized Solution

Define the G.P. Terms

Forming the A.P.

First A.P. Condition

Simplify First Equation

Factorize First Equation

Second A.P. Condition

Simplify Second Equation

Factorize Second Equation

Solve for Common Ratio $r$

Calculate First Term $a$

The Product Expression

Substitute and Calculate

Final Division

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Shift to Arithmetic

The Algebraic Bridge

The Elegant Cancellation

The Final Victory

Similar Questions

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Let $a$ and $b$ be two distinct positive real numbers. Let $1 1^{t h}$ term of a GP, whose first term is $a$ and third term is $b$ , is equal to $p^{t h}$ term of another GP, whose first term is $a$ and fifth term is $b$ . Then $p$ is equal to

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

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Select Answer:

Visualized Solution

Define the G.P. Terms

Forming the A.P.

First A.P. Condition

Simplify First Equation

Factorize First Equation

Second A.P. Condition

Simplify Second Equation

Factorize Second Equation

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

Calculate First Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

The Product Expression

Substitute and Calculate

Final Division

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Shift to Arithmetic

The Algebraic Bridge

The Elegant Cancellation

The Final Victory

Similar Questions

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If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:

Solve for Common Ratio $r$

Calculate First Term $a$

For the two positive numbers $a, b$ , if $a, b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a}, 10$ and $\frac{1}{b}$ are in an arithmetic progression, then, $16 a + 12 b$ is equal to