Let Tr be the rth term of an A.P. If for some m, Tm = (1/25), T25 = (1/20) and 20 = 125 Tr = 13 then 5m = m2m Tr is…

Q: Let Tr be the rth term of an A.P. If for some m, Tm = (1/25), T25 = (1/20) and 20 = 125 Tr = 13 then 5m = m2m Tr is equal to:

### Related Formula Standard Arithmetic Progression summation template: ### Core Logic Given structural constraints: ### Step 1: Finding Parameters and Indices Substituting a = (1/500) back into a + 24d = (1/20) gives d = (1/500). Using the formula for Tm: ### Step 2: Computing the Target Segment Sum For m = 20, the target expression becomes: Evaluating the values gives exactly 126. ### Pattern Recognition When a = d, the expressions simplify directly to basic multiples of the index position (Tn

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

Let Tᵣ$T_r$ be the rth$r^{\text{th}}$ term of an A.P. If for some m, Tm = (1)/(25), T₂₅ = (1)/(20)$T_m = \frac{1}{25}, T_{25} = \frac{1}{20}$ and 20Σr = 1²⁵ Tᵣ = 13$20\sum_{r = 1}^{25} T_r = 13$ then 5mΣr = m2m Tᵣ$5m\sum_{r = m}^{2m} T_r$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding Parameters and Indices

Step 2: Computing the Target Segment Sum

Pattern Recognition

Chapter Mix

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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Let $Tᵣ$ $T_r$ be the $rth$ $r^{\text{th}}$ term of an A.P. If for some m, $Tm = (1)/(25), T₂₅ = (1)/(20)$ and $20Σr = 1²⁵ Tᵣ = 13$ $20\sum_{r = 1}^{25} T_r = 13$ then $5mΣr = m2m Tᵣ$ $5m\sum_{r = m}^{2m} T_r$ is equal to: