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…

Year	2026	2025	2024	Total
Questions	17	24	14	55

Q24 jee_main_2026_22_january_evening AP and GP Relations

Suppose

a, b, c

are in A.P. and

a², 2b², c²

are in G.P. If

a < b < c

and

a + b + c = 1

, then

9(a² + b² + c²)

is equal to ____.

Numerical Answer. Answer: 9 to 9

Solution

Related Formula

For A.P.: $a = b-d, c = b+d$ $a = b-d, c = b+d$. For G.P.: $(2b²)² = a² c² 4b⁴ = a² c²$ .

Core Logic

Given $a + b + c = 1 (b-d) + b + (b+d) = 1 3b = 1 b = (1)/(3)$ . Using G.P. condition:

4b⁴ = [(b-d)(b+d)]² = (b² - d²)²

4 ((1)/(81)) = ((1)/(9) - d²)² (1)/(9) - d² = \pm (2)/(9)

Since $a < b < c$ $a < b < c$, $d > 0$ $d > 0$. Taking $(1)/(9) - d² = -(2)/(9) d² = (1)/(3) d = 1\sqrt(3)$ .

Step 1: Compute Required Expression

a² + b² + c² = (b-d)² + b² + (b+d)² = 3b² + 2d²

= 3((1)/(9)) + 2((1)/(3)) = (1)/(3) + (2)/(3) = 1

9(a² + b² + c²) = 9(1) = 9

Pattern Recognition

Set $a=b-d, c=b+d$ $a=b-d, c=b+d$ to eliminate terms early using $a+b+c=1$ $a+b+c=1$.

Chapter Mix

Class 11 Maths: Sequences and Series

Q20 jee_main_2026_23_january_evening Sum of n Terms

Let

Σk=1ⁿ ak = α n² + β n

. If

a₁₀ = 59

and

a₆ = 7a₁

then

α + β

is equal to

A. $12$ $12$
B. $3$ $3$
C. $5$ $5$
D. $7$ $7$

Solution

Related Formula

The $n$ $n$-th term of a sequence is $aₙ = Sₙ - Sₙ₋₁$ $a_n = S_n - S_{n-1}$.

Core Logic

We have $Sₙ = α n² + β n$ $S_n = \alpha n^2 + \beta n$.

aₙ = (α n² + β n) - (α(n-1)² + β(n-1))

aₙ = α(n² - (n² - 2n + 1)) + β(n - n + 1)

aₙ = α(2n - 1) + β

This shows the sequence is an Arithmetic Progression with common difference $2α$ $2\alpha$.

Step 1: Using Given Conditions

Condition 1: $a₁₀ = 59$ $a_{10} = 59$

a₁₀ = α(2(10) - 1) + β = 19α + β = 59

Condition 2: $a₆ = 7a₁$ $a_6 = 7a_1$

a₆ = α(2(6) - 1) + β = 11α + β

a₁ = α(2(1) - 1) + β = α + β

Substitute into the condition:

11α + β = 7(α + β)

11α + β = 7α + 7β

4α = 6β 2α = 3β

Step 2: Solving for Alpha and Beta

Substitute $2α = 3β β = (2)/(3)α$ into the first equation:

19α + (2)/(3)α = 59

(57α + 2α)/(3) = 59

(59α)/(3) = 59 α = 3

If $α = 3$ $\alpha = 3$, then $β = (2)/(3)(3) = 2$ $\beta = \frac{2}{3}(3) = 2$. Therefore, $α + β = 3 + 2 = 5$ $\alpha + \beta = 3 + 2 = 5$.

Pattern Recognition

Any sum of $n$ $n$ terms that is purely a quadratic function in $n$ $n$ describes an AP. You can instantly map the common difference to $2α$ $2\alpha$ and the first term to $α + β$ $\alpha + \beta$.

Chapter Mix

Class 11 Maths: Sequences and Series

Q4 jee_main_2026_24_january_morning Geometric Progressions and Product Sequences

Let

729, 81, 9, 1,

be a sequence and

Pₙ

denote the product of the first

n

terms of this sequence. If

2Σn=1⁴⁰(Pₙ)(1)/(n) = (3^α - 1)/(3^β)

and

(α, β) = 1

, then

α + β

is equal to

A. $73$ $73$
B. $74$ $74$
C. $75$ $75$
D. $76$ $76$

Solution

Related Formula

Pₙ = a₁ \cdot a₂ aₙ

Sum of GP: $Sₙ = a ( (1 - rⁿ)/(1 - r) )$

Core Logic

The sequence is $3⁶, 3⁴, 3², 3⁰,$ $3^6, 3^4, 3^2, 3^0, \dots$ General term $ak = 36 - 2(k-1) = 38 - 2k$ $a_k = 3^{6 - 2(k-1)} = 3^{8 - 2k}$.

Pₙ = 3⁶ \cdot 3⁴ \cdot 3² 38 - 2n

Pₙ = 36 + 4 + 2 + + (8 - 2n)

The exponent is an AP with first term $6$ $6$, common difference $-2$ $-2$, up to $n$ $n$ terms. Sum = $(n)/(2) [2(6) + (n-1)(-2)] = (n)/(2) [14 - 2n] = n(7 - n)$ .

Pₙ = 3n(7 - n)

Step 1: Simplifying the Expression

(Pₙ)(1)/(n) = (3n(7 - n))(1)/(n) = 37 - n

Now we evaluate the sum:

Σn=1⁴⁰ (Pₙ)(1)/(n) = Σn=1⁴⁰ 37 - n

= 3⁶ + 3⁵ + 3⁴ + (40 terms)

Step 2: Geometric Series Sum

This is a GP with first term $a = 3⁶$ $a = 3^6$, common ratio $r = (1)/(3)$ $r = \frac{1}{3}$, and $40$ $40$ terms.

S₄₀ = 3⁶ [ 1 - (1/3)⁴⁰1 - 1/3 ] = 3⁶ ( 1 - 3⁻⁴⁰ )2/3 = (3⁷)/(2) ( 3⁴⁰ - 13⁴⁰ )

= 3⁴⁰ - 12 \times 3³³

Step 3: Comparing with Given Form

Given expression:

2 Σn=1⁴⁰ (Pₙ)(1)/(n) = 2 \times 3⁴⁰ - 12 \times 3³³ = 3⁴⁰ - 13³³

Comparing with $(3^α - 1)/(3^β)$ $\frac{3^\alpha - 1}{3^\beta}$, we get:

α = 40, β = 33

Check $(40, 33) = 1$ $\gcd(40, 33) = 1$. (True) Therefore, $α + β = 40 + 33 = 73$ $\alpha + \beta = 40 + 33 = 73$.

Pattern Recognition

Product sequences of exponent-based terms collapse neatly when finding the $n$ $n$-th root, reducing complex product operations into standard Geometric Progressions.

Chapter Mix

Class 11 Maths: Sequences and Series

Q15 jee_main_2026_24_january_morning Arithmetic Progression Properties

Consider an A.P.:

a₁, a₂, , aₙ; a₁ > 0

. If

a₂ - a₁ = (-3)/(4), aₙ = (1)/(4) a₁

, and

Σi=1ⁿ aᵢ = (525)/(2)

, then

Σi=1¹⁷ aᵢ

is equal to

A. $476$ $476$
B. $952$ $952$
C. $238$ $238$
D. $136$ $136$

Solution

Related Formula

Sₙ = (n)/(2)(a₁ + aₙ)

aₙ = a₁ + (n - 1)d

Core Logic

Given common difference $d = -3/4$ $d = -3/4$. $(n)/(2)(a₁ + (a₁)/(4)) = (525)/(2)$ $(n)/(2) \cdot (5a₁)/(4) = (525)/(2) \Rightarrow (5na₁)/(4) = 525 \Rightarrow n a₁ = 420$

Step 1: Finding Parameters

Use $n$ $n$-th term formula:

aₙ = a₁ + (n-1)d \Rightarrow (a₁)/(4) = a₁ + (n-1)((-3)/(4))

(-3)/(4)a₁ = (-3)/(4)(n-1) \Rightarrow a₁ = n-1

Substitute into $n a₁ = 420$ $n a_1 = 420$: $n(n-1) = 420$ $n(n-1) = 420$

n² - n - 420 = 0 \Rightarrow (n-21)(n+20) = 0

Since $n > 0$ $n > 0$, $n = 21$ $n = 21$. This means $a₁ = 20$ $a_1 = 20$.

Step 2: Evaluating S17

Σi=1¹⁷ aᵢ = S₁₇ = (17)/(2) [2a₁ + 16d]

= (17)/(2) [ 2(20) + 16((-3)/(4)) ]

= (17)/(2) [40 - 12] = (17)/(2) (28) = 17 \times 14 = 238

Pattern Recognition

Dual AP conditions defining sum and last term invariably reduce to a quadratic in $n$ $n$. Using $Sₙ = (n)/(2)(a + l)$ $S_n = \frac{n}{2}(a + l)$ avoids messy expansion of $(n-1)d$ $(n-1)d$ prematurely.

Chapter Mix

Class 11 Maths: Sequences and Series

Q5 jee_main_2026_24_january_evening Arithmetic Progression Properties

Let

α₁, α₂, α₃, α₄

be an A.P. of four terms such that each term of the A.P. and its common difference

l

are integers. If

α₁ + α₂ + α₃ + α₄ = 48

and

α₁α₂α₃α₄ + l⁴ = 361

then the largest term of the A.P. is equal to

A. $27$ $27$
B. $24$ $24$
C. $21$ $21$
D. $23$ $23$

Solution

Related Formula

Standard 4-term A.P. representation: a-3d, a-d, a+d, a+3d

Common difference here is 2d = l

Core Logic

Let the terms $α₁, α₂, α₃, α₄$ be $a - 3d, a - d, a + d, a + 3d$ $a - 3d, a - d, a + d, a + 3d$.

Sum of terms = $48$ $48$:

4a = 48 a = 12

Step 1: Product Condition

We are given $α₁α₂α₃α₄ + l⁴ = 361$ . Since the common difference is $l = 2d$ $l = 2d$, $l⁴ = 16d⁴$ $l^4 = 16d^4$.

(a - 3d)(a - d)(a + d)(a + 3d) + 16d⁴ = 361

(a² - 9d²)(a² - d²) + 16d⁴ = 361

Substitute $a = 12$ $a = 12$ ( $a² = 144$ $a^2 = 144$):

(144 - 9d²)(144 - d²) + 16d⁴ = 361

20736 - 144d² - 1296d² + 9d⁴ + 16d⁴ = 361

25d⁴ - 1440d² + 20736 = 361

Step 2: Factoring the Quartic

Notice that $25d⁴ - 1440d² + 20736 = (5d² - 144)²$ .

(5d² - 144)² = 361 = 19²

5d² - 144 = \pm 19

Case 1: $5d² = 144 + 19 = 163 d² = (163)/(5)$ (Rejected, as $d$ $d$ would not yield integer common differences).

Case 2: $5d² = 144 - 19 = 125 d² = 25 d = \pm 5$ .

This gives common difference $l = 2d = 10$ $l = 2d = 10$, which is an integer. Thus, $d=5$ $d=5$ is valid.

Step 3: Calculating the Largest Term

The largest term of the A.P. is $a + 3d$ $a + 3d$:

α₄ = 12 + 3(5) = 12 + 15 = 27

Pattern Recognition

The product of four symmetrically spaced A.P. terms $(a-3d)(a-d)(a+d)(a+3d) + (2d)⁴$ $(a-3d)(a-d)(a+d)(a+3d) + (2d)^4$ strictly simplifies to a perfect square: $(a² - 5d²)²$ $(a^2 - 5d^2)^2$. Knowing this identity instantly skips the heavy expansion algebra.

Chapter Mix

Class 11 Maths: Sequences 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)$\mathbb{R} - (a, b)$, then a² + b²$a^2 + b^2$ is equal to ____.

Solution & Explanation

Core Logic

Step 1: Finding the Sum

Step 2: Range of S

Step 3: Calculating Final Answer

Chapter Mix

Reference Study Guides

More Sequences and Series Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Compute Required Expression

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Using Given Conditions

Step 2: Solving for Alpha and Beta

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplifying the Expression

Step 2: Geometric Series Sum

Step 3: Comparing with Given Form

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Parameters

Step 2: Evaluating S17

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Product Condition

Step 2: Factoring the Quartic

Step 3: Calculating the Largest Term

Pattern Recognition

Chapter Mix

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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 $R - (a, b)$ $\mathbb{R} - (a, b)$, then $a² + b²$ $a^2 + b^2$ is equal to ____.