JEE Main 2026 (28 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $R - (a, b)$ , then $a^{2} + b^{2}$ is equal to ......... .

Q: Wait, why do we choose the terms as $\frac{A}{r}, A, Ar$ instead of $a, ar, ar^2$?

That is a fantastic question! While both methods are mathematically valid, choosing terms symmetrically is a classic JEE strategy. When you multiply $\frac{A}{r} \cdot A \cdot Ar$, the $r$ terms cancel out perfectly, leaving you with $A^3 = 27$. If you had used $a, ar, ar^2$, you would get $a^3 r^3 = 27$, which is $(ar)^3 = 27$, leading to $ar = 3$. You would then have to express everything in terms of $a$ and $r$, which is much more algebraically tedious. Symmetry is your best friend in competitive exams!

Q: Why do we need to consider $r < 0$? Does the problem say $r$ is positive?

The problem does not state that $r$ is positive, and in the world of JEE Advanced, if a condition isn't explicitly given, you must assume the most general case. A common ratio $r$ can absolutely be negative. If you ignore the negative values, you will miss half the possible range of the sum $S$, which would lead to an incorrect excluded interval. Always check the boundaries and the signs of your variables!

Q: How did you get the excluded interval $(-3, 9)$? I thought the range was just $S \ge 9$.

You are halfway there! For $r > 0$, the AM-GM inequality gives us $r + \frac{1}{r} \ge 2$, which leads to $S \ge 9$. But for $r 0$. Then $r + \frac{1}{r} = -(k + \frac{1}{k})$. Since $k + \frac{1}{k} \ge 2$, it follows that $-(k + \frac{1}{k}) \le -2$. Substituting this into our sum formula, we get $S \le 3(-2 + 1) = -3$. So, $S$ can be in $(-\infty, -3]$ or $[9, \infty)$. The values between $-3$ and $9$ are never reached, hence the excluded interval is $(-3, 9)$.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark: Visualize the symmetry of the Geometric Progression. Instead of using the standard

a, a r, a r^{2}

, consider the terms as

\frac{A}{r}, A, A r

. This choice is a strategic masterstroke because when you multiply these three terms, the common ratio

r

vanishes, leaving you with a simple cubic equation for the middle term

A

Hint 2

Hint 2: The Arsenal: You must employ the AM-GM inequality, but with a twist. Recall that for any positive real number

x

x + \frac{1}{x} \geq 2

. However, the problem does not restrict the common ratio

r

to be positive. You must analyze the behavior of the function

f (r) = r + \frac{1}{r}

for both positive and negative values of

r

separately.

Hint 3

Hint 3: The Execution: Start by expressing the sum

S

S = 3 (r + \frac{1}{r} + 1)

. Now, determine the range of the expression

r + \frac{1}{r}

for

r > 0

and

r < 0

. Once you have the two disjoint intervals for

S

, identify the excluded interval

(a, b)

that lies between them.

Visualized Solution

Defining the G.P. Terms

Let the first three terms of the G.P. be $\frac{A}{r}, A, A r$
where $A$ is the middle term and $r$ is the common ratio ( $r \neq = 0$ )

Using the Product Condition

Given: Product of terms $= 27$
$(\frac{A}{r}) \cdot (A) \cdot (A r) = 27$

Finding the Middle Term

$A^{3} = 27 ⟹ A = 3$
The terms are $\frac{3}{r}, 3, 3 r$

Expressing the Sum $S$

Let the sum of the first three terms be $S$
$S = \frac{3}{r} + 3 + 3 r$
$S = 3 (r + \frac{1}{r} + 1)$

Analyzing $r + \frac{1}{r}$ for $r > 0$

For $r > 0$ , by AM-GM inequality: $r + \frac{1}{r} \geq 2$
Equality holds at $r = 1$

Range of $S$ for $r > 0$

Substitute $r + \frac{1}{r} \geq 2$ into $S$
$S \geq 3 (2 + 1) ⟹ S \geq 9$

Analyzing $r + \frac{1}{r}$ for $r < 0$

For $r < 0$ , let $r = - k$ where $k > 0$
$r + \frac{1}{r} = - (k + \frac{1}{k}) \leq - 2$
Equality holds at $r = - 1$

Range of $S$ for $r < 0$

Substitute $r + \frac{1}{r} \leq - 2$ into $S$
$S \leq 3 (- 2 + 1) ⟹ S \leq - 3$

Finding the Excluded Interval

The possible values of $S$ are $(- \infty, - 3] \cup [9, \infty)$
The excluded interval is $(- 3, 9)$
Given excluded interval is $(a, b)$ , so $a = - 3$ and $b = 9$

Final Calculation: $a^{2} + b^{2}$

We need to find $a^{2} + b^{2}$
$a^{2} + b^{2} = (- 3)^{2} + 9^{2}$
$a^{2} + b^{2} = 9 + 81 = 90$

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

When dealing with the product of three terms in a Geometric Progression (G.P.), the choice of variables is critical for efficiency. Instead of the standard

a, a r, a r^{2}

, we utilize the symmetric form:

\frac{A}{r}, A, A r

This choice is powerful because the common ratio

r

cancels out during multiplication, simplifying the algebra significantly. Given that the product of these terms is

27

, we write:

(\frac{A}{r}) \cdot (A) \cdot (A r) = 27

The

r

terms cancel out, leaving us with

A^{3} = 27

. Taking the cube root, we find the middle term:

A = 3

The Heart of the Problem

The Sum Expression

With the middle term identified, our G.P. terms are

\frac{3}{r}, 3, 3 r

. The sum

S

is defined as:

S = \frac{3}{r} + 3 + 3 r

Factoring out the constant

3

, we obtain:

S = 3 (r + \frac{1}{r} + 1)

The behavior of the sum

S

is entirely dependent on the function

f (r) = r + \frac{1}{r}

. To determine the range of

S

, we must analyze the range of this function across all possible real values of

r

(where $r eq 0$).

The Trap of the Negative Ratio

Many students incorrectly apply the AM-GM inequality only for

r > 0

, concluding that

r + \frac{1}{r} \geq 2

. This leads to the incomplete result

S \geq 3 (2 + 1) = 9

However, we must account for negative values of

r

. If

r < 0

, let

r = - k

where

k > 0

. Then:

r + \frac{1}{r} = - (k + \frac{1}{k})

Since

k + \frac{1}{k} \geq 2

for all

k > 0

, it follows that

r + \frac{1}{r} \leq - 2

for all negative

r

. Substituting this into our expression for

S

S \leq 3 (- 2 + 1) = - 3

The Final Synthesis

We have determined that the sum

S

exists in the intervals

(- \infty, - 3]

and

[9, \infty)

. Consequently, the values strictly between

- 3

and

9

are unattainable.

The excluded interval is

(- 3, 9)

. Given this interval is defined as

(a, b)

, we identify:

a = - 3, b = 9

The final calculation is:

a^{2} + b^{2} = (- 3)^{2} + 9^{2} = 9 + 81 = 90

The final result is 90.

Common Traps & FAQs

Wait, why do we choose the terms as

\frac{A}{r}, A, A r

instead of

a, a r, a r^{2}

That is a fantastic question! While both methods are mathematically valid, choosing terms symmetrically is a classic JEE strategy. When you multiply

\frac{A}{r} \cdot A \cdot A r

, the

r

terms cancel out perfectly, leaving you with

A^{3} = 27

. If you had used

a, a r, a r^{2}

, you would get

a^{3} r^{3} = 27

, which is

(a r)^{3} = 27

, leading to

a r = 3

. You would then have to express everything in terms of

a

and

r

, which is much more algebraically tedious. Symmetry is your best friend in competitive exams!

Why do we need to consider

r < 0

? Does the problem say

r

is positive?

The problem does not state that

r

is positive, and in the world of JEE Advanced, if a condition isn't explicitly given, you must assume the most general case. A common ratio

r

can absolutely be negative. If you ignore the negative values, you will miss half the possible range of the sum

S

, which would lead to an incorrect excluded interval. Always check the boundaries and the signs of your variables!

How did you get the excluded interval

(- 3, 9)

? I thought the range was just

S \geq 9

You are halfway there! For

r > 0

, the AM-GM inequality gives us

r + \frac{1}{r} \geq 2

, which leads to

S \geq 9

. But for

r < 0

, let

r = - k

where

k > 0

. Then

r + \frac{1}{r} = - (k + \frac{1}{k})

. Since

k + \frac{1}{k} \geq 2

, it follows that

- (k + \frac{1}{k}) \leq - 2

. Substituting this into our sum formula, we get

S \leq 3 (- 2 + 1) = - 3

. So,

S

can be in

(- \infty, - 3]

[9, \infty)

. The values between

- 3

and

9

are never reached, hence the excluded interval is

(- 3, 9)

Animated Solution for Mathematics - Sequence and Series: In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $R - (a, b)$ , then $a^{2} + b^{2}$ is equal to ......... .

Enter Numerical Value:

Visualized Solution

Defining the G.P. Terms

Using the Product Condition

Finding the Middle Term

Expressing the Sum $S$

Analyzing $r + \frac{1}{r}$ for $r > 0$

Range of $S$ for $r > 0$

Analyzing $r + \frac{1}{r}$ for $r < 0$

Range of $S$ for $r < 0$

Finding the Excluded Interval

Final Calculation: $a^{2} + b^{2}$

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

Analyzing the Setup

The Heart of the Problem

The Trap of the Negative Ratio

The Final Synthesis

Similar Questions

Let $a_{1}, a_{2}, a_{3}, \dots$ be a GP of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then $a_{1} a_{9} + a_{2} a_{4} a_{9} + a_{5} + a_{7}$ is equal to

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and.eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

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

The sum of first four terms of a geometric progression (G.P.) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$ . If the product of first three terms of the G.P. is 1, and the third term is $α$ , then $2 α$ is

If $a_{1} (> 0), a_{2}, a_{3}, a_{4}, a_{5}$ are in a G.P., $a_{2} + a_{4} = 2 a_{3} + 1$ and $3 a_{2} + a_{3} = 2 a_{4}$ , then $a_{2} + a_{4} + 2 a_{5}$ is equal to ______.

Let $729, 81, 9, 1, \dots$ be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence. If $2 \sum_{n = 1}^{40} (P_{n})^{1/ n} = \frac{3 ^{α} - 1}{3 ^{β}}$ and $g cd (α, β) = 1$ , then $α + β$ is equal to

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 of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :

Let the first term $a$ and the common ratio $r$ of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to

If a, b, c are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. such that $a < b < c$ and $a + b + c = \frac{3}{4}$ , then the value of a is :-

Enter Numerical Value:

Visualized Solution

Defining the G.P. Terms

Using the Product Condition

Finding the Middle Term

Expressing the Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r+r1​ for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r>0

Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r>0

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r+r1​ for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r<0

Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r<0

Finding the Excluded Interval

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2+b2

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

Analyzing the Setup

The Heart of the Problem

The Trap of the Negative Ratio

The Final Synthesis

Similar Questions

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and.eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a1​(>0),a2​,a3​,a4​,a5​ are in a G.P., a2​+a4​=2a3​+1 and 3a2​+a3​=2a4​, then a2​+a4​+2a5​ is equal to ______.

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a, b, c are in A.P. and a2,b2,c2 are in G.P. such that a<b<c and a+b+c=43​, then the value of a is :-

Expressing the Sum $S$

Analyzing $r + \frac{1}{r}$ for $r > 0$

Range of $S$ for $r > 0$

Analyzing $r + \frac{1}{r}$ for $r < 0$

Range of $S$ for $r < 0$

Final Calculation: $a^{2} + b^{2}$

If $a_{1} (> 0), a_{2}, a_{3}, a_{4}, a_{5}$ are in a G.P., $a_{2} + a_{4} = 2 a_{3} + 1$ and $3 a_{2} + a_{3} = 2 a_{4}$ , then $a_{2} + a_{4} + 2 a_{5}$ is equal to ______.

If a, b, c are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. such that $a < b < c$ and $a + b + c = \frac{3}{4}$ , then the value of a is :-