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

Q59 jee_main_2025_24_jan_evening Arithmetic Progression Sum

In an arithmetic progression, if

S₄₀=1030

and

S₁₂=57,

then

S₃₀-S₁₀

is equal to:

A. $510$ $510$
B. $515$ $515$
C. $525$ $525$
D. $505$ $505$

Solution

Related Formula

Sum of first $n$ $n$ terms of an Arithmetic Progression:

Sₙ = (n)/(2)[2a + (n-1)d]

Core Logic

Set up linear expressions for the given sums :

S₄₀ = (40)/(2)[2a + 39d] = 1030 \Rightarrow 2a + 39d = 51.5

S₁₂ = (12)/(2)[2a + 11d] = 57 \Rightarrow 2a + 11d = 9.5

Step 1: Solve for $a$ $a$ and $d$ $d$

Subtract the second equation from the first :

(2a + 39d) - (2a + 11d) = 51.5 - 9.5

28d = 42 \Rightarrow d = (42)/(28) = (3)/(2) = 1.5

Substitute $d = 1.5$ $d = 1.5$ back to find $a$ $a$:

2a + 11(1.5) = 9.5 \Rightarrow 2a + 16.5 = 9.5 \Rightarrow 2a = -7 \Rightarrow a = -3.5

Step 2: Evaluate $S₃₀ - S₁₀$ $S_{30} - S_{10}$

Write out the formula for the target subtraction :

S₃₀ - S₁₀ = (30)/(2)[2a + 29d] - (10)/(2)[2a + 9d]

= 15(2a + 29d) - 5(2a + 9d) = 30a + 435d - 10a - 45d = 20a + 390d

Substitute the values of $a$ $a$ and $d$ $d$ :

= 20(-3.5) + 390(1.5) = -70 + 585 = 515

Pattern Recognition

Notice that $S₃₀ - S₁₀$ $S_{30} - S_{10}$ represents the sum of terms from $T₁₁$ $T_{11}$ to $T₃₀$ $T_{30}$, which can also be formulated as $20 \times A20.5$ $20 \times A_{20.5}$, saving algebraic steps if calculated symmetrically.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q60 jee_main_2025_24_jan_evening Arithmetico-Geometric Progression

If

7=5+(1)/(7)(5+α)+ 17²(5+2α)+ 17³(5+3α)+ \infty

, then the value of

α

is:

A. $1$ $1$
B. $(6)/(7)$ $\frac{6}{7}$
C. $6$ $6$
D. $(1)/(7)$ $\frac{1}{7}$

Solution

Related Formula

Sum of an infinite geometric progression:

S\infty = (a)/(1-r) for |r| < 1

Core Logic

The given expression is an infinite Arithmetico-Geometric Progression (AGP) :

S = 5 + (5+α)/(7) + (5+2α)/(7²) + (5+3α)/(7³) + \infty

Step 1: Shift and Subtract

Multiply the equation by the common ratio $(1)/(7)$ $\frac{1}{7}$ and shift it by one position :

(1)/(7)S = (5)/(7) + (5+α)/(7²) + (5+2α)/(7³) + \infty

Subtract this from the original equation:

S - (1)/(7)S = 5 + ((5+α-5)/(7)) + ((5+2α-(5+α))/(7²)) +

(6)/(7)S = 5 + (α)/(7) + (α)/(7²) + (α)/(7³) +

Step 2: Sum the Infinite Geometric Series

Apply the infinite GP formula to the terms involving $α$ $\alpha$ :

(6)/(7)S = 5 + (α)/(7)((1)/(1 - (1)/(7))) = 5 + (α)/(7)((7)/(6)) = 5 + (α)/(6)

Given that $S = 7$ $S = 7$ :

(6)/(7)(7) = 5 + (α)/(6) \Rightarrow 6 = 5 + (α)/(6)

1 = (α)/(6) \Rightarrow α = 6

Pattern Recognition

Standard trick for infinite AGPs: Multiply by the common ratio $r$ $r$, shift, and subtract to condense the arithmetic progression component into a straightforward infinite geometric progression.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q54 jee_main_2025_24_jan_morning Sum to n terms of Special Series

Let

Sₙ = (1)/(2) + (1)/(6) + (1)/(12) + (1)/(20) +

up to

n

terms. If the sum of the first six terms of an A.P. with first term

-p

and common difference

p

is

2026S₂₀₂₅

, then the absolute difference between

20th

and

15th

terms of the A.P. is :

A. $25$ $25$
B. $90$ $90$
C. $20$ $20$
D. $45$ $45$

Solution

Related Formula

The general term for the provided series is:

Tk = (1)/(k(k+1)) = (1)/(k) - (1)/(k+1)

This sets up a standard telescoping summation sequence.

Core Logic

Express the sum $S₂₀₂₅$ $S_{2025}$ via telescoping fractions:

S₂₀₂₅ = Σk=1²⁰²⁵ ( (1)/(k) - (1)/(k+1) ) = (1 - (1)/(2)) + ((1)/(2) - (1)/(3)) + + ((1)/(2025) - (1)/(2026))

S₂₀₂₅ = 1 - (1)/(2026) = (2025)/(2026)

Step 1: Compute the boundary expression value

Substitute $S₂₀₂₅$ $S_{2025}$ into the expression value:

2026 \cdot S₂₀₂₅ = \sqrt(2026 \cdot (2025)/(2026)) = \sqrt(2025) = 45

Step 2: Apply Arithmetic Progression Summation

The sum of the first 6 terms of the A.P. with $a = -p$ $a = -p$ and $d = p$ $d = p$ is equal to 45:

Σ₆ = (6)/(2) [2a + (6-1)d] = 45

3 [2(-p) + 5p] = 45

3 [3p] = 45 9p = 45 p = 5

Step 3: Calculate target absolute term difference

The absolute difference between the $20th$ $20^{\text{th}}$ and $15th$ $15^{\text{th}}$ terms of any A.P. depends strictly on the common difference:

|A₂₀ - A₁₅| = |(a + 19p) - (a + 14p)| = 5p

$5p = 5(5) = 25$ $5p = 5(5) = 25$

Pattern Recognition

The series sequence $(1)/(2) + (1)/(6) + (1)/(12) +$ is the well-known telescoping series $Σ (1)/(n(n+1))$ $\sum \frac{1}{n(n+1)}$. Its sum to $n$ $n$ terms is identically given by $(n)/(n+1)$ $\frac{n}{n+1}$ without requiring manual re-derivation.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q63 jee_main_2025_28_jan_evening Telescopic Series Summation

For positive integers n, if

4aₙ=(n²+5n+6)

and

Sₙ=Σk=1ⁿ( 1ak)

then the value of 507

S₂₀₂₅

is:

A. $540$ $540$
B. $1350$ $1350$
C. $675$ $675$
D. $135$ $135$

Solution

Related Formula

Telescopic series decomposition via method of differences:

(1)/((k+2)(k+3)) = (1)/(k+2) - (1)/(k+3)

Core Logic

Given:

aₙ = (n²+5n+6)/(4) = ((n+2)(n+3))/(4)

Therefore, the reciprocal term is:

(1)/(ak) = (4)/((k+2)(k+3)) = 4 [ (1)/(k+2) - (1)/(k+3) ]

Step 1: Compute the Partial Sum

Sₙ = Σk=1ⁿ (1)/(ak) = 4 Σk=1ⁿ ( (1)/(k+2) - (1)/(k+3) )

Expanding the sum terms:

Sₙ = 4 [ ((1)/(3) - (1)/(4)) + ((1)/(4) - (1)/(5)) + + ((1)/(n+2) - (1)/(n+3)) ]

All intermediate terms cancel out:

Sₙ = 4 [ (1)/(3) - (1)/(n+3) ] = 4 [ (n+3 - 3)/(3(n+3)) ] = (4n)/(3(n+3))

Step 2: Calculate for n = 2025

For $n = 2025$ $n = 2025$:

S₂₀₂₅ = (4 \times 2025)/(3 \times (2025 + 3)) = (4 \times 2025)/(3 \times 2028)

We need to find $507 \times S₂₀₂₅$ $507 \times S_{2025}$:

507 \times S₂₀₂₅ = 507 \times (4 \times 2025)/(3 \times 2028)

Notice that $2028 = 4 \times 507$ $2028 = 4 \times 507$:

507 \times S₂₀₂₅ = 507 \times (4 \times 2025)/(3 \times (4 \times 507)) = (2025)/(3) = 675

Pattern Recognition

Always look for arithmetic factor groupings at the end of large number sequence questions in JEE. Here recognizing $2028 = 4 \times 507$ $2028 = 4 \times 507$ avoids large multi-digit multiplication.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q73 jee_main_2025_28_jan_evening Arithmetic Progression Applications

The interior angles of a polygon with n sides, are in an A.P. with common difference

6°

If the largest interior angle of the polygon is

219°

, then n is equal to

Numerical Answer. Answer: 20 to 20

Solution

Related Formula

Sum of interior angles of an $n$ $n$-sided polygon:

Sₙ = (n - 2) \times 180^°

Sum of an Arithmetic Progression:

Sₙ = (n)/(2) [ 2a + (n-1)d ]

Core Logic

The angles form an AP with common difference $d = 6^°$ $d = 6^\circ$. The largest angle is the last term: $Tₙ = 219^°$ $T_n = 219^\circ$.

a + (n-1)6 = 219 a = 219 - 6n + 6 = 225 - 6n

Step 1: Set up the sum equation

Equating the two forms for the sum of angles:

(n)/(2) [ 2a + (n-1)6 ] = (n - 2) \times 180

Substitute $a = 225 - 6n$ $a = 225 - 6n$:

(n)/(2) [ 2(225 - 6n) + 6n - 6 ] = 180n - 360

(n)/(2) [ 450 - 12n + 6n - 6 ] = 180n - 360

(n)/(2) [ 444 - 6n ] = 180n - 360

n(222 - 3n) = 180n - 360

222n - 3n² = 180n - 360

3n² - 42n - 360 = 0

Step 2: Solve the Quadratic Equation

Divide by 3:

n² - 14n - 120 = 0

(n - 20)(n + 6) = 0

Since number of sides $n$ $n$ must be positive, $n = 20$ $n = 20$.

Pattern Recognition

Always remember that any interior angle of a convex polygon must be less than $180^°$ $180^\circ$. Let's check the smallest angle for $n=20$ $n=20$: $a = 225 - 120 = 105^°$ $a = 225 - 120 = 105^\circ$, which is completely valid.

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 8

Solution

Related Formula

Core Logic

Step 1: Solve for a$a$ and d$d$

Step 2: Evaluate S₃₀ - S₁₀$S_{30} - S_{10}$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Shift and Subtract

Step 2: Sum the Infinite Geometric Series

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute the boundary expression value

Step 2: Apply Arithmetic Progression Summation

Step 3: Calculate target absolute term difference

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute the Partial Sum

Step 2: Calculate for n = 2025

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set up the sum equation

Step 2: Solve the Quadratic Equation

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:

Step 1: Solve for $a$ $a$ and $d$ $d$

Step 2: Evaluate $S₃₀ - S₁₀$ $S_{30} - S_{10}$