JEE Main 2026 (24 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: 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

Q: Why did we reject the root $k = 130.4$?

The problem explicitly states that the common difference $l$ is an integer. Since $k = l^2$, $k$ must be a perfect square. $130.4$ is not a perfect square, so it is physically impossible for this specific A.P. Always check your constraints against the problem statement!

Q: Does it matter if $l$ is positive or negative?

Not for the set of numbers! If $l=10$, the sequence is $-3, 7, 17, 27$. If $l=-10$, the sequence is $27, 17, 7, -3$. The set of numbers remains identical, and the largest term remains $27$ regardless of the direction of the progression.

Select Answer:

Visualized Solution

Choosing Symmetric Terms for A.P.

Let the four terms of the A.P. be:
$(a - 3 d), (a - d), (a + d), (a + 3 d)$
Common difference $l = 2 d$
All terms and $l$ are integers.

Using the Sum Condition

Sum of terms: $(a - 3 d) + (a - d) + (a + d) + (a + 3 d) = 48$
Simplifying: $4 a = 48$
Result: $a = 12$

Expressing Terms using $l$

Since $l = 2 d$ , we have $d = \frac{l}{2}$
Substitute $a = 12$ and $d = \frac{l}{2}$ into the terms:
Terms: $(12 - \frac{3 l}{2}), (12 - \frac{l}{2}), (12 + \frac{l}{2}), (12 + \frac{3 l}{2})$

Setting up the Product Equation

Product condition: $a_{1} a_{2} a_{3} a_{4} + l^{4} = 361$
$(12 - \frac{3 l}{2}) (12 - \frac{l}{2}) (12 + \frac{l}{2}) (12 + \frac{3 l}{2}) + l^{4} = 361$
Group terms: $[(12 - \frac{3 l}{2}) (12 + \frac{3 l}{2})] \cdot [(12 - \frac{l}{2}) (12 + \frac{l}{2})] + l^{4} = 361$

Simplifying the Product

Using $(x - y) (x + y) = x^{2} - y^{2}$ :
$(144 - \frac{9 l ^{2}}{4}) (144 - \frac{l ^{2}}{4}) + l^{4} = 361$

Substitution and Expansion

Let $k = l^{2}$
$(144 - \frac{9 k}{4}) (144 - \frac{k}{4}) + k^{2} = 361$
Expanding: $20736 - 36 k - 324 k + \frac{9 k ^{2}}{16} + k^{2} = 361$
$20736 - 360 k + \frac{25 k ^{2}}{16} = 361$

Solving the Quadratic Equation

Rearranging: $\frac{25 k ^{2}}{16} - 360 k + 20375 = 0$
Multiply by $16$ : $25 k^{2} - 5760 k + 326000 = 0$
Solving for $k$ : $k = 100$ or $k = 130.4$

Finding the Common Difference $l$

Since $l$ is an integer, $k = l^{2}$ must be a perfect square.
$k = 100$ is a perfect square, while $130.4$ is not.
$l^{2} = 100 ⟹ l = \pm 10$

Final Terms and Largest Value

Take $l = 10$ , then $d = 5$ and $a = 12$ .
Terms: $(12 - 15), (12 - 5), (12 + 5), (12 + 15)$
Terms: $- 3, 7, 17, 27$
The largest term is $27$ .

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

Analyzing the Setup

The standard approach to an Arithmetic Progression (A.P.) is to define terms as

a, a + d, a + 2 d, a + 3 d

. However, for an even number of terms, we can utilize symmetry to save time.

By choosing our terms as

a - 3 d, a - d, a + d, a + 3 d

, we engineer a cancellation. When we sum these terms, the

d

terms vanish, leaving us with:

4 a = 48

This immediately yields $a = 12$ . This choice of variables acts as a strategic coordinate system for the problem.

The Algebraic Heavy Lifting

We face the product condition:

(a_{1} a_{2} a_{3} a_{4}) + l^{4} = 361

. With

a = 12

and

l = 2 d

(implying

d = l /2

), our terms become:

(12 - \frac{3 l}{2}), (12 - \frac{l}{2}), (12 + \frac{l}{2}), (12 + \frac{3 l}{2})

To avoid drowning in algebra, we group the terms by pairing the first with the fourth and the second with the third. Applying the difference of squares identity,

(x - y) (x + y) = x^{2} - y^{2}

, the product transforms into:

(144 - \frac{9 l ^{2}}{4}) (144 - \frac{l ^{2}}{4}) + l^{4} = 361

Taming the Quadratic Beast

To simplify, we introduce the substitution

k = l^{2}

. The equation becomes:

(144 - \frac{9 k}{4}) (144 - \frac{k}{4}) + k^{2} = 361

Expanding this expression with precision, we arrive at:

20736 - 360 k + \frac{9 k ^{2}}{16} + k^{2} = 361

Rearranging this into a standard quadratic form, we obtain:

25 k^{2} - 5760 k + 326000 = 0

Solving for

k

, we find two potential values:

k = 100

k = 130.4

The Final Victory

The constraint of the problem acts as our filter. Since

l

must be an integer,

k

(which is

l^{2}

) must be a perfect square.

Because

130.4

is not a perfect square, we discard it. Thus,

l^{2} = 100

, which gives us

l = \pm 10

With

l = 10

, we find

d = 5

. Substituting these back into our symmetric terms, we get the sequence:

- 3, 7, 17, 27

The largest term is $27$ . You have successfully navigated the trap of brute-force calculation by observing the underlying structure of the problem.

Common Traps & FAQs

Why did we choose

a - 3 d

and

a - d

instead of the standard

a, a + d, a + 2 d, a + 3 d

It is all about reducing the number of variables. By centering the A.P. around the middle, the sum becomes

4 a

instead of

4 a + 6 d

. This is a classic JEE time-saver that eliminates the need to solve a system of equations for

a

and

d

simultaneously.

Why did we reject the root

k = 130.4

The problem explicitly states that the common difference

l

is an integer. Since

k = l^{2}

k

must be a perfect square.

130.4

is not a perfect square, so it is physically impossible for this specific A.P. Always check your constraints against the problem statement!

Does it matter if

l

is positive or negative?

Not for the set of numbers! If

l = 10

, the sequence is

- 3, 7, 17, 27

. If

l = - 10

, the sequence is

27, 17, 7, - 3

. The set of numbers remains identical, and the largest term remains

27

regardless of the direction of the progression.

Select Answer:

Visualized Solution

Choosing Symmetric Terms for A.P.

Using the Sum Condition

Expressing Terms using $l$

Setting up the Product Equation

Simplifying the Product

Substitution and Expansion

Solving the Quadratic Equation

Finding the Common Difference $l$

Final Terms and Largest Value

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

Analyzing the Setup

The Algebraic Heavy Lifting

Taming the Quadratic Beast

The Final Victory

Similar Questions

The common difference of the A.P.: $a_{1}, a_{2}, \dots, a_{m}$ is 13 more than the common difference of the A.P.: $b_{1}, b_{2}, \dots, b_{n}$ . If $b_{31} = - 277, b_{43} = - 385$ and $a_{78} = 327$ , then $a_{1}$ is equal to

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 :

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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:

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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 $a_{1} = 8, a_{2}, a_{3}, \dots, a_{n}$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is

Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ________.

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

Select Answer:

Visualized Solution

Choosing Symmetric Terms for A.P.

Using the Sum Condition

Expressing Terms using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l

Setting up the Product Equation

Simplifying the Product

Substitution and Expansion

Solving the Quadratic Equation

Finding the Common Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l

Final Terms and Largest Value

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

Analyzing the Setup

The Algebraic Heavy Lifting

Taming the Quadratic Beast

The Final Victory

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In an A.P., the sixth terms a6​=2. If the a1​a4​a5​ is the greatest, then the common difference of the A.P., is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​=8,a2​,a3​,…,an​ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is

Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Tr​ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m,n,m=n,Tm​=1/n and Tn​=1/m, then a−d equals

Expressing Terms using $l$

Finding the Common Difference $l$

In an A.P., the sixth terms $a_{6} = 2$ . If the $a_{1} a_{4} a_{5}$ is the greatest, then the common difference of the A.P., is equal to

Let $a_{1} = 8, a_{2}, a_{3}, \dots, a_{n}$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals