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

Q70 jee_main_2025_03_april_evening Special Series

The sum

1 + (1+3)/(2!) + (1+3+5)/(3!) + (1+3+5+7)/(4!) +

is equal to

A. $6e$ $6e$
B. $4e$ $4e$
C. $3e$ $3e$
D. $2e$ $2e$

Solution

Related Formula

Sum of first $r$ $r$ odd natural numbers:

Σk=1r (2k-1) = r²

Exponential series expansion:

Σr=0\infty (1)/(r!) = e

Core Logic

Let's find the general $r$ $r$-th term of the series:

Tᵣ = (1 + 3 + 5 + + (2r-1))/(r!) = (r²)/(r!) = (r)/((r-1)!)

Step 1: Expressing term in terms of sum limits

Let's write $r = (r-1) + 1$ $r = (r-1) + 1$:

Tᵣ = (r-1+1)/((r-1)!) = (1)/((r-2)!) + (1)/((r-1)!)

Our infinite sum is:

S = Σr=1\infty Tᵣ = Σr=2\infty (1)/((r-2)!) + Σr=1\infty (1)/((r-1)!)

Both sums are standard representations of the exponential expansion.

Step 2: Summing the parts

First part: $Σr=2\infty (1)/((r-2)!) = 1 + (1)/(1!) + (1)/(2!) + = e$
Second part: $Σr=1\infty (1)/((r-1)!) = 1 + (1)/(1!) + (1)/(2!) + = e$

Total Sum S = e + e = 2e

Pattern Recognition

The general term containing $r²$ $r^2$ in summation with factorials converges to $2e$ $2e$. Remember the shortcut: $Σ (r²)/(r!) = 2e$ $\sum \frac{r^2}{r!} = 2e$, $Σ (r³)/(r!) = 5e$ $\sum \frac{r^3}{r!} = 5e$. It is extremely useful to memorize these common limits.

Chapter Mix

Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Limits, Continuity and Differentiability

Q63 jee_main_2025_07_april_morning Arithmetico-Geometric Progression

Let

x₁, x₂, x₃, x₄

be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from

x₁, x₂, x₃, x₄

then the resulting numbers are in an arithmetic progression. Then the value of

(1)/(24) (x₁ x₂ x₃ x₄)

is:

A. $72$ $72$
B. $18$ $18$
C. $36$ $36$
D. $216$ $216$

Solution

Related Formula

For a geometric progression, the terms can be set as $a, ar, ar², ar³$ $a, ar, ar^2, ar^3$. For three terms $A, B, C$ $A, B, C$ to be in arithmetic progression, they must satisfy: $2B = A + C$ $2B = A + C$

Core Logic

Let the elements be $x₁ = a, x₂ = ar, x₃ = ar², x₄ = ar³$ . After the specified subtractions, the sequence becomes:

a - 2, ar - 7, ar² - 9, ar³ - 5

Since this sequence is in AP, we form two separate common difference linear linkages:

2(ar - 7) = (a - 2) + (ar² - 9) 2ar - 14 = ar² + a - 11 ar² - 2ar + a + 3 = 0 (1)

2(ar² - 9) = (ar - 7) + (ar³ - 5) 2ar² - 18 = ar³ + ar - 12 ar³ - 2ar² + ar + 6 = 0 (2)

Step 1: Solve the Simultaneous Polynomials

Multiply equation (1) by $r$ $r$:

ar³ - 2ar² + ar + 3r = 0 (3)

Subtract equation (3) from equation (2):

(ar³ - 2ar² + ar + 6) - (ar³ - 2ar² + ar + 3r) = 0

6 - 3r = 0 3r = 6 r = 2

Substitute $r = 2$ $r = 2$ back into equation (1):

a(2)² - 2a(2) + a + 3 = 0

4a - 4a + a + 3 = 0 a = -3

Step 2: Find the Continuous Product Value

The continuous product term is:

x₁x₂x₃x₄ = a \cdot ar \cdot ar² \cdot ar³ = a⁴ r⁶

x₁x₂x₃x₄ = (-3)⁴ \cdot (2)⁶ = 81 \times 64 = 5184

Now divide by 24 as required:

(1)/(24)(5184) = 216

Pattern Recognition

Notice that multiplying the first AP condition equation by $r$ $r$ perfectly mimics the structure of the second condition equation except for the absolute scalar value, allowing direct elimination of all polynomial variable indices simultaneously.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q jee_main_2025_08_april_evening Infinite Series

If

(1)/(1⁴) + (1)/(2⁴) + (1)/(3⁴) + \infty = (π⁴)/(90)

, and

(1)/(1⁴) + (1)/(3⁴) + (1)/(5⁴) + \infty = α

(1)/(2⁴) + (1)/(4⁴) + (1)/(6⁴) + \infty = β

then

(α)/(β)

is equal to

A. $23$ $23$
B. $18$ $18$
C. $15$ $15$
D. $14$ $14$

Solution

Related Formula

Total Sum = α + β

Core Logic

Factor out common fractions from the even terms component ( $β$ $\beta$) to represent it as a scalar multiple of the universal sum sequence.

Step 1: Simplify the Even Terms Series

β = (1)/(2⁴) + (1)/(4⁴) + (1)/(6⁴) + = (1)/(2⁴) ( (1)/(1⁴) + (1)/(2⁴) + (1)/(3⁴) + )

β = (1)/(16) ( (π⁴)/(90) )

Step 2: Express Alpha by Remainder Deduction

Since total sum equals $α + β$ $\alpha + \beta$:

α = Total Sum - β = (π⁴)/(90) - (1)/(16) ( (π⁴)/(90) ) = (15)/(16) ( (π⁴)/(90) )

Step 3: Compute the Relative Ratio

$(α)/(β) = ((15)/(16) ( (π⁴)/(90) ))/((1)/(16) ( (π⁴)/(90) )) = 15$

Pattern Recognition

For alternating p-series powers like $Σ n-p$ $\sum n^{-p}$, the even component fractions always condense via factor steps to $2-p \cdot Stotal$ $2^{-p} \cdot S_{\text{total}}$, decoupling power values cleanly from final simple scalar quotients.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q57 jee_main_2025_29_jan_evening Infinite Geometric Progression

Let

S = N 0

. Define a relation \mathbf{R} from

S

to

R

by:

R = (x, y): ₑ y = x ₑ ((2)/(5)), x in S, y in R .

Then, the sum of all the elements in the range of

R

is equal to

A. $(3)/(2)$ $\frac{3}{2}$
B. $(5)/(3)$ $\frac{5}{3}$
C. $(10)/(9)$ $\frac{10}{9}$
D. $(5)/(2)$ $\frac{5}{2}$

Solution

Related Formula

Sum of an infinite geometric progression with $|r| < 1$ $|r| < 1$:

S\infty = (a)/(1 - r)

Core Logic

From the definition of the relation:

ₑ y = x ₑ((2)/(5)) ₑ y = ₑ((2)/(5))^x y = ((2)/(5))^x

Infinite Geometric Progression diagram for Q57 - JEE Main 2025 Evening

Since $x in S = 0, 1, 2, 3,$ $x \in S = \{0, 1, 2, 3, \dots\}$, the output values of $y$ $y$ represent elements of the range.

Step 1: Compute Infinite Sum

Generating elements by plugging in values of $x$ $x$: For $x = 0 y = 1$ $x = 0 \implies y = 1$ For $x = 1 y = (2)/(5)$ $x = 1 \implies y = \frac{2}{5}$ For $x = 2 y = ((2)/(5))²$

Sum of elements in the range:

Sum = 1 + ((2)/(5))¹ + ((2)/(5))² + = (1)/(1 - (2)/(5)) = (5)/(3)

Pattern Recognition

Convert log equations into standard exponential equations right away. A variable index belonging to whole numbers indicates an infinite GP summation scenario.

Chapter Mix

Class 11 Mathematics: Sequences and Series Class 11 Mathematics: Relations and Functions

Q73 jee_main_2025_29_jan_evening Arithmetic Progression Properties

Let

a₁, a₂, \dots, a₂₀₂₄

be an Arithmetic Progression such that

a₁ + (a₅ + a₁₀ + a₁₅ + \dots + a₂₀₂₀) + a₂₀₂₄ = 2233

. Then

a₁ + a₂ + a₃ + \dots + a₂₀₂₄

is equal to

Numerical Answer. Answer: 11132 to 11132

Solution

Related Formula

Symmetry identity rule inside Arithmetic Progressions:

ak + an-k+1 = a₁ + aₙ

Core Logic

Group matching paired steps equidistant from sequence boundary ends:

a₁ + a₂₀₂₄ = a₅ + a₂₀₂₀ = a₁₀ + a₂₀₁₅ =

The sequence of inner indices follows an AP tracking loop:

5, 10, 15, , 2020

Calculate internal block element count $N$ $N$:

2020 = 5 + (N-1)5 2015 = 5(N-1) N - 1 = 403 N = 404 terms

Step 1: Simplify Expression Equations

Since the inner sequence contains 404 terms, they form exactly 202 symmetrical pairs. Adding $a₁$ $a_1$ and $a₂₀₂₄$ $a_{2024}$ introduces one more pair, resulting in 203 identical sum blocks:

203(a₁ + a₂₀₂₄) = 2233 a₁ + a₂₀₂₄ = (2233)/(203) = 11

Step 2: Evaluate the Total Sum

Using the standard AP sum formula:

S₂₀₂₄ = (2024)/(2)(a₁ + a₂₀₂₄) = 1012 \times 11 = 11132

Pattern Recognition

Progressions possess natural positional balance. Grouping symmetrical index pairs ( $ak + an-k+1$ $a_k + a_{n-k+1}$) allows factoring out variable steps right away.

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 5

Solution

Related Formula

Core Logic

Step 1: Expressing term in terms of sum limits

Step 2: Summing the parts

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve the Simultaneous Polynomials

Step 2: Find the Continuous Product Value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify the Even Terms Series

Step 2: Express Alpha by Remainder Deduction

Step 3: Compute the Relative Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Infinite Sum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify Expression Equations

Step 2: Evaluate the Total Sum

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: