For some n 10, let the coefficients of the 5^ th, 6^ th, and 7^ t… — Solution | JEE Mathematics

MCQMediumJEE 2025Binomial Expansion

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:

A
$70$
B
$35$
C
$20$
D
$10$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The coefficients of the $5^{\text{th}}$ , $6^{\text{th}}$ , and $7^{\text{th}}$ terms in $(1+x)^{n+4}$ are in A.P.

Find: The largest coefficient in the expansion.

In the expansion of $(1+x)^{n+4}$ , the coefficient of the $(k+1)^{\text{th}}$ term is $\binom{n+4}{k}$ . Hence the three coefficients are

\binom{n+4}{4},\; \binom{n+4}{5},\; \binom{n+4}{6}

and since they are in A.P.,

2\binom{n+4}{5}=\binom{n+4}{4}+\binom{n+4}{6}

Detailed Working from the Extracted Solution

Using the A.P. condition,

\binom{n+4}{5}=\frac{\binom{n+4}{4}+\binom{n+4}{6}}{2}

2\cdot {}^{n+4}C_5 = {}^{n+4}C_4 + {}^{n+4}C_6

The extracted solution's second approach simplifies this condition and states

n=3

Therefore,

(1+x)^{n+4}=(1+x)^7

Use the Middle Coefficient Idea

For $(1+x)^7$ , the greatest binomial coefficients are the middle ones. Since the power is odd, the middle two coefficients are equal:

{}^7C_3 = {}^7C_4 = \frac{7!}{3!4!}=35

Therefore, the largest coefficient is $35$ , so the correct option is B.

Note: Another extracted approach on the page mentions $n=6$ and $\binom{10}{5}=252$ , but it also explicitly says that this does not align with the given options. The page's resolved answer and the consistent second approach both give $35$ .

Common mistakes

Using the coefficient of the $k^{\text{th}}$ term as $\binom{n+4}{k}$ instead of $\binom{n+4}{k-1}$ . This shifts all three required coefficients by one place. Always remember that the $(r+1)^{\text{th}}$ term has coefficient $\binom{n+4}{r}$ .
Applying the A.P. condition incorrectly as $a-b=b-c$ with wrong sign handling. The safer form is $2b=a+c$ for three terms in arithmetic progression.
Assuming the largest coefficient is at a unique middle term without checking whether the exponent is odd or even. For an odd power such as $7$ , the two middle coefficients are equal and both are largest.

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Answer

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