JEE Advanced 1983

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: Let $n$ be positive integer. If the coefficients of 2nd, 3rd, and 4th terms in the expansion of $(1 + x)^{n}$ are in A.P., then the value of $n$ is .........

Q: Wait, why is the coefficient of the 4th term ${}^nC_3$ and not ${}^nC_4$?

This is a classic trap! In the binomial expansion of $(1+x)^n$, the general term is $T_{r+1} = {}^nC_r x^r$. Notice the index is $r+1$. So, for the 4th term, we set $r+1 = 4$, which means $r = 3$. Thus, the coefficient is ${}^nC_3$. Always be careful with the $r+1$ shift!

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Visualize the binomial expansion of

(1 + x)^{n}

. The coefficients are not just numbers; they are the binomial coefficients

^{n} C_{r}

. For the 2nd, 3rd, and 4th terms, identify the correct values of

r

in the general term

T_{r + 1} =^{n} C_{r} x^{r}

. Remember, the 2nd term corresponds to

r = 1

, the 3rd to

r = 2

, and the 4th to

r = 3

Hint 2

The Arsenal: You are given that these three coefficients are in an Arithmetic Progression (A.P.). Recall the fundamental property of any three numbers

a, b, c

in A.P.: the middle term is the arithmetic mean of the other two, which means

2 b = a + c

. Apply this to your identified coefficients.

Hint 3

The Execution: Substitute the expressions for the coefficients into the A.P. condition:

2 (^{n} C_{2}) =^{n} C_{1} +^{n} C_{3}

. Expand these combinations using the formula

^{n} C_{r} = \frac{n !}{r ! ( n - r )!}

and solve the resulting polynomial equation for

n

. Don't forget to check the constraint that

n

must be a positive integer and large enough for the 4th term to exist.

Visualized Solution

$T_{r + 1}$ and Coefficients

Expansion: $(1 + x)^{n}$
General term: $T_{r + 1} =^{n} C_{r} x^{r}$
Coefficient of 2nd term ( $T_{2}$ ): $^{n} C_{1}$
Coefficient of 3rd term ( $T_{3}$ ): $^{n} C_{2}$
Coefficient of 4th term ( $T_{4}$ ): $^{n} C_{3}$

A.P. Condition

Given: $^{n} C_{1},^{n} C_{2},^{n} C_{3}$ are in A.P.
Condition for A.P.: $2 \times (Middle Term) = First Term + Third Term$
Equation: $2 (^{n} C_{2}) =^{n} C_{1} +^{n} C_{3}$

Expand Combinations

Formula: $^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$
$^{n} C_{1} = n$
$^{n} C_{2} = \frac{n ( n - 1 )}{2}$
$^{n} C_{3} = \frac{n ( n - 1 ) ( n - 2 )}{6}$

Substitute into Equation

Substitute the expanded forms into the A.P. equation:
$2 (\frac{n ( n - 1 )}{2}) = n + \frac{n ( n - 1 ) ( n - 2 )}{6}$

Simplify and Divide by $n$

Cancel the $2$ on the left side:
$n (n - 1) = n + \frac{n ( n - 1 ) ( n - 2 )}{6}$
Since $n \geq 3$ , $n \neq = 0$ . Divide the entire equation by $n$ :
$(n - 1) = 1 + \frac{( n - 1 ) ( n - 2 )}{6}$

Clear the Denominator

Multiply the entire equation by $6$ to remove the fraction:
$6 (n - 1) = 6 (1) + (n - 1) (n - 2)$
$6 n - 6 = 6 + (n - 1) (n - 2)$

Expand the Product

Expand the binomial product $(n - 1) (n - 2)$ :
$(n - 1) (n - 2) = n^{2} - 2 n - n + 2 = n^{2} - 3 n + 2$
Substitute back into the equation:
$6 n - 6 = 6 + n^{2} - 3 n + 2$

Form Quadratic Equation

Rearrange all terms to one side:
$n^{2} - 3 n - 6 n + 6 + 2 + 6 = 0$
$n^{2} - 9 n + 14 = 0$

Factorize

Find factors of $14$ that sum to $- 9$ (which are $- 7$ and $- 2$ ):
$n^{2} - 7 n - 2 n + 14 = 0$
$n (n - 7) - 2 (n - 7) = 0$
$(n - 7) (n - 2) = 0$

Solve for $n$

Set each factor to zero:
$n - 7 = 0 \Rightarrow n = 7$
$n - 2 = 0 \Rightarrow n = 2$

Check Constraints

The problem involves a 4th term ( $T_{4}$ ).
For $T_{4}$ to exist in $(1 + x)^{n}$ , we must have $n \geq 3$ .
Therefore, $n = 2$ is rejected.
Final Answer: $n = 7$

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Imagine you are standing at the threshold of a binomial expansion, looking at the expression

(1 + x)^{n}

. The coefficients are the heartbeat of this expansion.

Using the general term formula

T_{r + 1} =^{n} C_{r} x^{r}

, we identify the key players: The 2nd term ( $r = 1$ ) gives us $^{n} C_{1}$ . The 3rd term (

r = 2

) gives us

^{n} C_{2}

. * The 4th term (

r = 3

) gives us

^{n} C_{3}

The A.P

Bridge

The problem states that these three coefficients are in an Arithmetic Progression. If three numbers

a, b, c

are in A.P., then

2 b = a + c

Applying this to our coefficients, we obtain the following equation:

2 (^{n} C_{2}) =^{n} C_{1} +^{n} C_{3}

This equation is the key that unlocks the value of

n

. It is a mathematical statement of balance.

The Algebraic Dance

We translate these combinations into the language of algebra using the standard definitions:

^{n} C_{1} = n,^{n} C_{2} = \frac{n ( n - 1 )}{2},^{n} C_{3} = \frac{n ( n - 1 ) ( n - 2 )}{6}

Substituting these into our A.P. equation, we get:

2 (\frac{n ( n - 1 )}{2}) = n + \frac{n ( n - 1 ) ( n - 2 )}{6}

The left side simplifies to

n (n - 1)

. Since we are dealing with a 4th term, we know

n \geq 3

, so $n eq 0$. We can safely divide the entire equation by

n

(n - 1) = 1 + \frac{( n - 1 ) ( n - 2 )}{6}

Clearing the denominator by multiplying by

6

, we transform the equation:

6 (n - 1) = 6 + (n - 1) (n - 2)

Expanding the right side yields:

6 n - 6 = 6 + n^{2} - 3 n + 2

Bringing all terms to one side, we arrive at the quadratic equation:

n^{2} - 9 n + 14 = 0

The Final Wisdom

Factorizing this quadratic, we look for two numbers that multiply to

14

and add to

- 9

. The numbers

- 7

and

- 2

fit perfectly.

Thus, we have:

(n - 7) (n - 2) = 0

This gives us two potential solutions:

n = 7

and

n = 2

. However, we must respect our constraints.

The problem requires a 4th term, which is impossible if

n = 2

. Therefore, we reject

n = 2

and embrace $n = 7$ as our final, triumphant answer.

Common Traps & FAQs

Wait, why is the coefficient of the 4th term

^{n} C_{3}

and not

^{n} C_{4}

This is a classic trap! In the binomial expansion of

(1 + x)^{n}

, the general term is

T_{r + 1} =^{n} C_{r} x^{r}

. Notice the index is

r + 1

. So, for the 4th term, we set

r + 1 = 4

, which means

r = 3

. Thus, the coefficient is

^{n} C_{3}

. Always be careful with the

r + 1

shift!

Why do we reject

n = 2

as a solution?

Great catch! The problem explicitly mentions the 4th term in the expansion. If

n = 2

, the expansion

(1 + x)^{2} = 1 + 2 x + x^{2}

only has 3 terms. For a 4th term to exist, we must have

n \geq 3

. Since

n = 2

violates this physical constraint of the expansion, we must discard it.

Can I just use the property that

^{n} C_{1},^{n} C_{2},^{n} C_{3}

are in A.P. without expanding the factorials?

While there are properties of binomial coefficients, expanding them into algebraic terms of

n

is the most robust and foolproof method for JEE. It turns a conceptual problem into a straightforward quadratic equation, minimizing the risk of missing a subtle constraint.

Animated Solution for Mathematics - Binomial Theorem: Let $n$ be positive integer. If the coefficients of 2nd, 3rd, and 4th terms in the expansion of $(1 + x)^{n}$ are in A.P., then the value of $n$ is .........

Enter Numerical Value:

Visualized Solution

$T_{r + 1}$ and Coefficients

A.P. Condition

Expand Combinations

Substitute into Equation

Simplify and Divide by $n$

Clear the Denominator

Expand the Product

Form Quadratic Equation

Factorize

Solve for $n$

Check Constraints

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The A.P

The Algebraic Dance

The Final Wisdom

Similar Questions

For some $n \neq = 10$ , let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of $(1 + x)^{n + 4}$ be in A.P. Then the largest coefficient in the expansion of $(1 + x)^{n + 4}$ is:

If the coefficients of $x^{4}, x^{5}$ and $x^{6}$ in the expansion of $(1 + x)^{n}$ are in the arithmetic progression, then the maximum value of $n$ is:

$r$ and $n$ are positive integers $r > 1, n > 2$ and coefficient of $(r + 2)^{t h}$ term and $3 r^{t h}$ term in the expansion of $(1 + x)^{2 n}$ are equal, then $n$ equals

The coefficients of three consecutive terms of $(1 + x)^{n + 5}$ are in the ratio $5 : 10 : 14$ . Then $n = .........$

If the coefficients of $r$ th, $(r + 1)$ th, and $(r + 2)$ th terms in the binomial expansion of $(1 + y)^{m}$ are in A.P., then $m$ and $r$ satisfy the equation

For a positive integer $n$ , $(1 + \frac{1}{x})^{n}$ is expanded in increasing powers of $x$ . If three consecutive coefficients in this expansion are in the ratio, $2 : 5 : 12$ , then $n$ is equal to

Let the coefficients of three consecutive terms in the binomial expansion of $(1 + 2 x)^{n}$ be in the ratio $2 : 5 : 8$ . Then the coefficient of the term, which is in the middle of these three terms, is ______.

If for some positive integer $n$ , the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n + 5}$ are in the ratio $5 : 10 : 14$ , then the largest coefficient in this expansion is :

Let $m$ be the smallest positive integer such that the coefficient of $x^{2}$ in the expansion of $(1 + x)^{2} + (1 + x)^{3} + ... + (1 + x)^{49} + (1 + m x)^{50}$ is $(3 n + 1)^{51} C_{3}$ for some positive integer $n$ . Then the value of $n$ is .........

The sum of the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n + 2}$ , which are in the ratio $1 : 3 : 5$ , is equal to

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr+1​ and Coefficients

A.P. Condition

Expand Combinations

Substitute into Equation

Simplify and Divide by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Clear the Denominator

Expand the Product

Form Quadratic Equation

Factorize

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

Check Constraints

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The A.P

The Algebraic Dance

The Final Wisdom

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the coefficients of x4,x5 and x6 in the expansion of (1+x)n are in the arithmetic progression, then the maximum value of n is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r and n are positive integers r>1,n>2 and coefficient of (r+2)th term and 3rth term in the expansion of (1+x)2n are equal, then n equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficients of three consecutive terms of (1+x)n+5 are in the ratio 5:10:14. Then n=.........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the coefficients of rth, (r+1)th, and (r+2)th terms in the binomial expansion of (1+y)m are in A.P., then m and r satisfy the equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a positive integer n, (1+x1​)n is expanded in increasing powers of x. If three consecutive coefficients in this expansion are in the ratio, 2:5:12, then n is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the coefficients of three consecutive terms in the binomial expansion of (1+x)n+2, which are in the ratio 1:3:5, is equal to

$T_{r + 1}$ and Coefficients

Simplify and Divide by $n$

Solve for $n$

If the coefficients of $x^{4}, x^{5}$ and $x^{6}$ in the expansion of $(1 + x)^{n}$ are in the arithmetic progression, then the maximum value of $n$ is:

$r$ and $n$ are positive integers $r > 1, n > 2$ and coefficient of $(r + 2)^{t h}$ term and $3 r^{t h}$ term in the expansion of $(1 + x)^{2 n}$ are equal, then $n$ equals

The coefficients of three consecutive terms of $(1 + x)^{n + 5}$ are in the ratio $5 : 10 : 14$ . Then $n = .........$

If the coefficients of $r$ th, $(r + 1)$ th, and $(r + 2)$ th terms in the binomial expansion of $(1 + y)^{m}$ are in A.P., then $m$ and $r$ satisfy the equation

For a positive integer $n$ , $(1 + \frac{1}{x})^{n}$ is expanded in increasing powers of $x$ . If three consecutive coefficients in this expansion are in the ratio, $2 : 5 : 12$ , then $n$ is equal to

The sum of the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n + 2}$ , which are in the ratio $1 : 3 : 5$ , is equal to