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

Q65 jee_main_2025_04_april_evening Arithmetic Progression

Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and q respectively. Let d and D be the common differences of AP's in A and B respectively such that

D = d + 3, ~d > 0

. If

p + qp - q = (19)/(5)

, then

p - q

is equal to

A. $600$ $600$
B. $450$ $450$
C. $630$ $630$
D. $540$ $540$

Solution

Core Logic

Let the 3 elements of set $A$ $A$ in A.P. be $a-d, a, a+d$ $a-d, a, a+d$. Their sum is $3a = 36 a = 12$ $3a = 36 \implies a = 12$. Their product is $p = a(a² - d²) = 12(144 - d²)$ $p = a(a^2 - d^2) = 12(144 - d^2)$.

Similarly, let the 3 elements of set $B$ $B$ be $b-D, b, b+D$ $b-D, b, b+D$. Their sum is $3b = 36 b = 12$ $3b = 36 \implies b = 12$. Their product is $q = b(b² - D²) = 12(144 - D²)$ $q = b(b^2 - D^2) = 12(144 - D^2)$.

Step 1: Using the Ratio Condition

We are given the relation:

(p + q)/(p - q) = (19)/(5)

Using componendo and dividendo:

(p)/(q) = (19 + 5)/(19 - 5) = (24)/(14) = (12)/(7)

Substitute the expression blocks for $p$ $p$ and $q$ $q$:

(12(144 - d²))/(12(144 - D²)) = (12)/(7) (144 - d²)/(144 - D²) = (12)/(7)

7(144 - d²) = 12(144 - D²)

Step 2: Substituting D in terms of d

We are given $D = d + 3$ $D = d + 3$:

7(144 - d²) = 12(144 - (d + 3)²)

1008 - 7d² = 12(144 - (d² + 6d + 9))

1008 - 7d² = 12(135 - d² - 6d) = 1620 - 12d² - 72d

5d² + 72d - 612 = 0

Solving this quadratic equation:

(d - 6)(5d + 102) = 0

Since $d > 0$ $d > 0$, we choose $d = 6$ $d = 6$. This implies $D = 6 + 3 = 9$ $D = 6 + 3 = 9$.

Step 3: Finding p - q

Now calculate the targeted metric:

p - q = 12(144 - d²) - 12(144 - D²) = 12(D² - d²)

p - q = 12(9² - 6²) = 12(81 - 36) = 12(45) = 540

Pattern Recognition

For 3-element symmetric AP sequences, choosing terms as $x-d, x, x+d$ $x-d, x, x+d$ ensures the sum isolates the middle term instantly ( $3x = S$ $3x = S$). This drastically drops algebraic variables from the start.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q53 jee_main_2025_04_april_morning Arithmetic Progression

Let

A = 1, 6, 11, 16,

and

B = 9, 16, 23, 30,

be the sets consisting of the first 2025 terms of two arithmetic progressions. Then

n(A B)

is

A. 3814
B. 4027
C. 3761
D. 4003

Solution

Related Formula

Set Principle of Inclusion-Exclusion:

n(A B) = n(A) + n(B) - n(A B)

Core Logic

Find the last terms of both progressions: For set $A$ $A$: $a₁ = 1, d₁ = 5 T₂₀₂₅ = 1 + (2025 - 1) \times 5 = 10121$ . For set $B$ $B$: $b₁ = 9, d₂ = 7 T₂₀₂₅ = 9 + (2025 - 1) \times 7 = 14177$ .

The intersection set $A B$ $A \cap B$ forms an AP with a common difference $d = LCM(5, 7) = 35$ $d = \text{LCM}(5, 7) = 35$. The first common term is $16$ $16$.

Step 1: Find Common Terms Count

The general term of the common AP must satisfy:

Tₙ = 16 + (n - 1) \times 35 \leq (10121, 14177) = 10121

(n - 1) \times 35 \leq 10105 n - 1 \leq 288.71 n = 289

Step 2: Total Distinct Terms

Apply the inclusion-exclusion principle:

n(A B) = 2025 + 2025 - 289 = 3761

Pattern Recognition

Common terms of two APs always generate a new AP whose common difference is the LCM of the individual common differences. Always verify the upper limit bound using the smaller of the two final values.

Chapter Mix

Class 11 Mathematics: Sequence and Series

Q61 jee_main_2025_04_april_morning Special Series

1 + 3 + 5² + 7 + 9² +

upto 40 terms is equal to

A. 43890
B. 41880
C. 33980
D. 40870

Solution

Related Formula

Summation Identities:

Σ r = (n(n+1))/(2), Σ r² = (n(n+1)(2n+1))/(6)

Core Logic

Split the 40-term series into two sub-series of 20 terms each: Series 1 (squared terms at positions 1, 3, 5... wait, positions are odd numbers whose base squares are odd): $1² + 5² + 9² +$ $1^2 + 5^2 + 9^2 + \dots$ upto 20 terms. General term $Tᵣ = (4r - 3)²$ $T_r = (4r - 3)^2$. Series 2 (linear terms at positions 2, 4, 6...): $3 + 7 + 11 +$ $3 + 7 + 11 + \dots$ upto 20 terms. General term $tᵣ = (4r - 1)$ $t_r = (4r - 1)$.

Step 1: Formulate Total Sigma Expression

Sum = Σr=1²⁰ [ (4r - 3)² + (4r - 1) ]

Sum = Σr=1²⁰ (16r² - 24r + 9 + 4r - 1) = Σr=1²⁰ (16r² - 20r + 8)

Sum = 16Σr=1²⁰ r² - 20Σr=1²⁰ r + 8Σr=1²⁰ 1

Step 2: Arithmetic Evaluation

Σr=1²⁰ r² = (20 \times 21 \times 41)/(6) = 2870

Σr=1²⁰ r = (20 \times 21)/(2) = 210

Sum = 16(2870) - 20(210) + 8(20) = 45920 - 4200 + 160 = 41880

Pattern Recognition

When dealing with interlaced series, pairing terms adjacent to each other simplifies the degree of general expressions into manageable standard summation polynomials.

Chapter Mix

Class 11 Mathematics: Sequence and Series

Q60 jee_main_2025_07_april_evening Arithmetic Progression

Let

aₙ

be the

nth

term of an A. P. If

Sₙ = a₁ + a₂ + a₃ + + aₙ = 700, a₆ = 7

and

S₇ = 7

, then

aₙ

is equal to:

A. $56$ $56$
B. $65$ $65$
C. $64$ $64$
D. $70$ $70$

Solution

Related Formula

Sum of first $n$ $n$ terms of an AP is given by:

Sₙ = (n)/(2)[2a + (n-1)d]

Core Logic

Given specifications:

$a₆ = 7 a + 5d = 7 (ii)$
$S₇ = 7 (7)/(2)(2a + 6d) = 7 a + 3d = 1 (iii)$

Subtracting (iii) from (ii):

2d = 6 d = 3

Substituting $d=3$ $d=3$ into (iii):

a + 3(3) = 1 a = -8

Step 1: Find n from Sn = 700

Substitute $a = -8$ $a = -8$ and $d = 3$ $d = 3$ into the equation for $Sₙ = 700$ $S_n = 700$:

700 = (n)/(2)[2(-8) + (n-1)3]

1400 = n[-16 + 3n - 3]

3n² - 19n - 1400 = 0

Factoring the quadratic equation:

(3n + 56)(n - 25) = 0

Since $n$ $n$ must be a positive integer, $n = 25$ $n = 25$.

Step 2: Determine standard term value

We need to find $a₂₅$ $a_{25}$ corresponding to index $n=25$ $n=25$:

a₂₅ = a + 24d

a₂₅ = -8 + 24(3) = -8 + 72 = 64

Pattern Recognition

When $Sₙ$ $S_n$ and specific terms are given, prioritize finding the first term $a$ $a$ and common difference $d$ $d$ through simple elimination headers before targeting the value of $n$ $n$.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q70 jee_main_2025_07_april_evening Geometric Progression

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 :

A. $745$ $745$
B. $755$ $755$
C. $750$ $750$
D. $757$ $757$

Solution

Related Formula

Sum of first $n$ $n$ terms of a Geometric Progression (GP) is:

Sₙ = (a(rⁿ - 1))/(r - 1)

Core Logic

Let the first term be $a$ $a$ and common ratio be $r$ $r$. Given:

$ar + ar³ + ar⁵ = 21 ar(1 + r² + r⁴) = 21 (1)$
$ar⁷ + ar⁹ + ar¹¹ = 15309 ar⁷(1 + r² + r⁴) = 15309 (2)$

Dividing equation (2) by equation (1):

(ar⁷)/(ar) = (15309)/(21) r⁶ = 729 r = 3

Step 1: Solve for a

Substitute $r = 3$ $r = 3$ into equation (1):

a(3)(1 + 9 + 81) = 21

3a(91) = 21 a = (7)/(91) = (1)/(13)

Step 2: Find Sum of 9 terms

Evaluating $S₉$ $S_9$:

S₉ = (a(r⁹ - 1))/(r - 1) = ((1)/(13)(3⁹ - 1))/(3 - 1) = (19683 - 1)/(26) = (19682)/(26) = 757

Pattern Recognition

Ratios of shifted groups of terms in a GP always cleanly isolate a simple power of the common ratio $r^k$ $r^k$ instantly.

Chapter Mix

Class 11 Mathematics: 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 7

Solution

Core Logic

Step 1: Using the Ratio Condition

Step 2: Substituting D in terms of d

Step 3: Finding p - q

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find Common Terms Count

Step 2: Total Distinct Terms

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Formulate Total Sigma Expression

Step 2: Arithmetic Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find n from Sn = 700

Step 2: Determine standard term value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for a

Step 2: Find Sum of 9 terms

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: