JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$ . Then the number of terms which are integers in the A.P. is :

Q: How do we know the odd terms form an A.P. with common difference $2d$?

Great observation! If the original sequence is $a_1, a_1+d, a_1+2d, a_1+3d, \dots$, then the odd terms are $a_1, a_1+2d, a_1+4d, \dots$. The difference between consecutive terms in this sub-sequence is $(a_1+2d) - a_1 = 2d$. It is a property of any sub-sequence taken at constant intervals from an A.P.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Do not rush to write the sum formula for the entire A.P. immediately. Instead, look at the difference between the sum of even-positioned terms and odd-positioned terms. Notice that

(a_{2} - a_{1}) + (a_{4} - a_{3}) + \dots + (a_{2 n} - a_{2 n - 1})

creates a beautiful, repeating pattern of the common difference

d

Hint 2

Hint 2: The Arsenal. You have two equations now: one involving

n d

(from the sum difference) and one involving

(2 n - 1) d

(from the last term condition). Treat

n

and

d

as your variables. Can you solve this system of two equations to isolate

d

and then

n

Hint 3

Hint 3: The Execution. Once you have

d = 1.5

and

n = 4

, you know the total number of terms is

2 n = 8

. Now, use the sum of odd terms formula:

S_{o dd} = \frac{n}{2} [2 a_{1} + (n - 1) (2 d)] = 24

to find the first term

a_{1}

. Once you have

a_{1}

and

d

, simply list the terms to count the integers.

Visualized Solution

Defining the A.P. Structure

Let the number of terms be $2 n$ .
Let the first term be $a_{1}$ and common difference be $d$ .
The terms are $a_{1}, a_{2}, a_{3}, \dots, a_{2 n}$ .

Setting up Sum Equations

Sum of odd terms: $a_{1} + a_{3} + \dots + a_{2 n - 1} = 24$
Sum of even terms: $a_{2} + a_{4} + \dots + a_{2 n} = 30$

Subtracting the Sums

Subtracting the odd sum from the even sum:
$(a_{2} - a_{1}) + (a_{4} - a_{3}) + \dots + (a_{2 n} - a_{2 n - 1}) = 30 - 24$

Finding the Relation $n d = 6$

Each pair $(a_{2 k} - a_{2 k - 1})$ equals the common difference $d$ .
Since there are $2 n$ terms, there are $n$ such pairs.
$d + d + \dots + d (n times) = 6$
$\Rightarrow n d = 6$

Using the Last Term Condition

Given: The last term exceeds the first by $\frac{21}{2}$ .
$a_{2 n} - a_{1} = \frac{21}{2}$
Formula for the last term: $a_{2 n} = a_{1} + (2 n - 1) d$

Substituting the Formula

Substitute $a_{2 n}$ into the condition:
$(a_{1} + (2 n - 1) d) - a_{1} = \frac{21}{2}$
$\Rightarrow (2 n - 1) d = \frac{21}{2}$

Expanding and Substituting $n d$

Expanding the bracket: $2 n d - d = \frac{21}{2}$
Substitute the known value $n d = 6$ :
$2 (6) - d = \frac{21}{2}$

Solving for $d$

$12 - d = 10.5$
$d = 12 - 10.5$
$d = 1.5 = \frac{3}{2}$

Solving for $n$

Using our relation $n d = 6$ :
$n (\frac{3}{2}) = 6$
$n = 6 \times \frac{2}{3} = 4$
Total number of terms $= 2 n = 8$

Finding the First Term $a_{1}$

We need $a_{1}$ to find the actual terms.
Use the sum of odd terms: $S_{o dd} = 24$
The odd terms form an A.P. with $n = 4$ terms, first term $a_{1}$ , and common difference $2 d$ .
Formula: $S_{o dd} = \frac{n}{2} [2 a_{1} + (n - 1) (2 d)]$

Substituting into Sum Formula

We know $n = 4$ and $d = \frac{3}{2}$ , so $2 d = 3$ .
Substitute these into the sum formula:
$\frac{4}{2} [2 a_{1} + (4 - 1) (3)] = 24$

Solving for $a_{1}$

Simplifying the equation:
$2 [2 a_{1} + 9] = 24$
$2 a_{1} + 9 = 12$
$2 a_{1} = 3 \Rightarrow a_{1} = \frac{3}{2} = 1.5$

Identifying Integer Terms

The A.P. starts at $1.5$ with a common difference of $1.5$ .
The $8$ terms are: $1.5, 3, 4.5, 6, 7.5, 9, 10.5, 12$
The integer terms are: $3, 6, 9, 12$
Total number of integer terms $= 4$

00:00 / 00:00

The Sigma Insight: Arithmetic Progression (A.P.)

Solution Diagram

Analyzing the Setup

We are given an Arithmetic Progression (A.P.) with

2 n

terms:

a_{1}, a_{2}, a_{3}, \dots, a_{2 n}

. The sum of the odd-positioned terms is

24

, and the sum of the even-positioned terms is

30

Instead of using the standard sum formula, we observe the symmetry of the sequence. We define the sum of odd terms as

S_{o dd}

and the sum of even terms as

S_{e v e n}

The Subtraction Trick

Consider the difference between the sum of even-positioned terms and odd-positioned terms:

S_{e v e n} - S_{o dd} = (a_{2} - a_{1}) + (a_{4} - a_{3}) + \dots + (a_{2 n} - a_{2 n - 1})

Each bracketed term represents the common difference

d

. Since there are

2 n

terms, there are exactly

n

such pairs.

n \times d = 30 - 24 = 6

This yields our first fundamental relation:

n d = 6

The Last Term Constraint

We are given that the last term exceeds the first by

\frac{21}{2}

. Mathematically, this is expressed as:

a_{2 n} - a_{1} = \frac{21}{2}

Using the general term formula

a_{2 n} = a_{1} + (2 n - 1) d

, we substitute to get:

(2 n - 1) d = \frac{21}{2}

Expanding this equation, we obtain:

2 n d - d = 10.5

Solving for Parameters

Substituting our known relation

n d = 6

into the equation above:

2 (6) - d = 10.5

12 - d = 10.5 \Rightarrow d = 1.5 = \frac{3}{2}

Now, we solve for

n

using

n d = 6

n (1.5) = 6 \Rightarrow n = 4

The total number of terms in the sequence is

2 n = 8

The Final Hunt

To find the first term

a_{1}

, we use the sum of the odd-positioned terms:

S_{o dd} = \frac{n}{2} [2 a_{1} + (n - 1) (2 d)] = 24

Substituting

n = 4

and

d = 1.5

(so

2 d = 3

\frac{4}{2} [2 a_{1} + (4 - 1) (3)] = 24

2 [2 a_{1} + 9] = 24

2 a_{1} + 9 = 12 \Rightarrow 2 a_{1} = 3 \Rightarrow a_{1} = 1.5

The sequence is

1.5, 3, 4.5, 6, 7.5, 9, 10.5, 12

. The integer terms in this sequence are

3, 6, 9, 12

There are exactly 4 integer terms in the sequence.

Common Traps & FAQs

Wait, why do we reject the negative root here? Or is there only one solution?

In this specific problem, the algebra leads to a linear system. When we solve

2 n d - d = 10.5

with

n d = 6

, we get a unique solution for

d

and

n

. There is no quadratic equation involved, so we don't have to worry about rejecting roots. The math is deterministic and elegant!

Why can't we just use the standard sum formula

S_{n} = \frac{n}{2} (2 a + (n - 1) d)

for both sums?

You absolutely can! However, that approach leads to a system of equations with

a_{1}

n

, and

d

that is algebraically cumbersome. The 'subtraction trick' we used is a classic JEE strategy to reduce complexity. It transforms a messy system into a clean, linear relationship between

n

and

d

. Always look for symmetry before brute-forcing formulas.

How do we know the odd terms form an A.P. with common difference

2 d

Great observation! If the original sequence is

a_{1}, a_{1} + d, a_{1} + 2 d, a_{1} + 3 d, \dots

, then the odd terms are

a_{1}, a_{1} + 2 d, a_{1} + 4 d, \dots

. The difference between consecutive terms in this sub-sequence is

(a_{1} + 2 d) - a_{1} = 2 d

. It is a property of any sub-sequence taken at constant intervals from an A.P.

Animated Solution for Mathematics - Sequence and Series: The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$ . Then the number of terms which are integers in the A.P. is :

Select Answer:

Visualized Solution

Defining the A.P. Structure

Setting up Sum Equations

Subtracting the Sums

Finding the Relation $n d = 6$

Using the Last Term Condition

Substituting the Formula

Expanding and Substituting $n d$

Solving for $d$

Solving for $n$

Finding the First Term $a_{1}$

Substituting into Sum Formula

Solving for $a_{1}$

Identifying Integer Terms

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Subtraction Trick

The Last Term Constraint

Solving for Parameters

The Final Hunt

Similar Questions

Suppose that the number of terms in an A.P. is 2k, $k \in N .$ If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to:

Let $a_{1}, a_{2}, a_{3}, a_{4}$ be an A.P. of four terms such that each term of the A.P. and its common difference $l$ are integers. If $a_{1} + a_{2} + a_{3} + a_{4} = 48$ and $a_{1} a_{2} a_{3} a_{4} + l^{4} = 361$ , then the largest term of the A.P. is equal to

Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its $1 1^{t h}$ term is :

If $a_{1}, a_{2}, a_{3}, \dots$ are in A.P. such that $a_{1} + a_{7} + a_{16} = 40$ , then the sum of the first 15 terms of this A.P. is :

Let $T_{r}$ be the rth term of an A.P. If for some m, $T_{m} = \frac{1}{25}$ and $T_{25} = \frac{1}{20}$ , then $5 m \sum_{r = m}^{2 m} T_{r}$ is equal to Note: $20 \sum_{r = 1}^{25} T_{r} = 13$ is given condition in Hindi version.

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots$ , then $n$ equals

Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6 : 11$ and the seventh term lies in between $130$ and $140$ , then the common difference of this A.P. is \dots.

Let $S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + ...$ upto n terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is $2026 S_{2025},$ then the absolute difference betwen $2 0^{t h}$ and $1 5^{t h}$ terms of the A.P. is equal to

Let $2^{n d}, 8^{t h}$ and $4 4^{t h}$ terms of a non-constant A.P. be respectively the $1^{s t}, 2^{n d}$ and $3^{r d}$ terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms is equal to-

If the $1 0^{t h}$ term of an A.P. is $\frac{1}{20}$ and its $2 0^{t h}$ term is $\frac{1}{10}$ , then the sum of its first 200 term is

Select Answer:

Visualized Solution

Defining the A.P. Structure

Setting up Sum Equations

Subtracting the Sums

Finding the Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nd=6

Using the Last Term Condition

Substituting the Formula

Expanding and Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nd

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Finding the First Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​

Substituting into Sum Formula

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​

Identifying Integer Terms

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Subtraction Trick

The Last Term Constraint

Solving for Parameters

The Final Hunt

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a1​,a2​,a3​,… are in A.P. such that a1​+a7​+a16​=40, then the sum of the first 15 terms of this A.P. is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the first 2n terms of the A.P. 2,5,8,… is equal to the sum of the first n terms of the A.P. 57,59,61,…, then n equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let 2nd,8th and 44th terms of a non-constant A.P. be respectively the 1st,2nd and 3rd terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms is equal to-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the 10th term of an A.P. is 201​ and its 20th term is 101​, then the sum of its first 200 term is

Finding the Relation $n d = 6$

Expanding and Substituting $n d$

Solving for $d$

Solving for $n$

Finding the First Term $a_{1}$

Solving for $a_{1}$

Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its $1 1^{t h}$ term is :

If $a_{1}, a_{2}, a_{3}, \dots$ are in A.P. such that $a_{1} + a_{7} + a_{16} = 40$ , then the sum of the first 15 terms of this A.P. is :

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots$ , then $n$ equals

Let $2^{n d}, 8^{t h}$ and $4 4^{t h}$ terms of a non-constant A.P. be respectively the $1^{s t}, 2^{n d}$ and $3^{r d}$ terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms is equal to-

If the $1 0^{t h}$ term of an A.P. is $\frac{1}{20}$ and its $2 0^{t h}$ term is $\frac{1}{10}$ , then the sum of its first 200 term is