For some $n \neq 10$, let the coefficients of the $5^{\text {th }}, 6^{\text {th }}$ and $7^{\text {th }}$ 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 :

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Binomial Theorem

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 :

Answer: (B)

The coefficients of the

5^{th}

6^{th}

7^{th}

terms of

(1 + x)^{n + 4}

(writing

N = n + 4

) are

(4 N), (5 N), (6 N)

.

Being in A.P.:

2 (5 N) = (4 N) + (6 N)

. Dividing through by

(5 N)

and using

\frac{( 4 N )}{( 5 N )} = \frac{5}{N - 4}

and

\frac{( 6 N )}{( 5 N )} = \frac{N - 5}{6}

2 = \frac{5}{N - 4} + \frac{N - 5}{6} .

Multiplying by

6 (N - 4)

12 (N - 4) = 30 + (N - 5) (N - 4) \Rightarrow 12 N - 48 = N^{2} - 9 N + 50 \Rightarrow N^{2} - 21 N + 98 = 0.

N = \frac{21 \pm 441 - 392}{2} = \frac{21 \pm 7}{2} = 14 or 7.

N = n + 4 = 14

gives

n = 10

, which is excluded; so

N = 7

, i.e.,

n = 3

.

The largest coefficient in

(1 + x)^{7}

is the middle binomial coefficient:

(3 7) = (4 7) = 35.

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option