Binomial (1+x)^(n+4): 5th, 6th, 7th Coefficients in AP, Find Largest Coefficient

JEE Main 2025 — 24 January, Morning Shift. Previous Year Question.

Problem

For some $n \neq 10$, consider the binomial expansion of $(1+x)^{n+4}$. If the coefficients of the 5th, 6th, and 7th terms in this expansion are in arithmetic progression, find the largest coefficient in the expansion of $(1+x)^{n+4}$.

Key insight. “Three terms in AP” always means “twice the middle term equals the sum of the outer two.” Writing that single equation using binomial coefficients — rather than trying to compute each one separately — turns the whole problem into one equation in $n$ that factors cleanly.

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Approach

Write the coefficients of the 5th, 6th, and 7th terms of $(1+x)^{n+4}$ as binomial coefficients, impose the AP condition $2\times(\text{middle}) = (\text{first})+(\text{last})$, and simplify using the ratio identities between consecutive binomial coefficients. This produces a quadratic in $n$, which — after excluding the disallowed root $n=10$ — pins down the exact expansion, whose largest coefficient is then just its middle binomial coefficient.

Solution

Step 1 — Write the three coefficients

Let $N = n+4$. The 5th, 6th, and 7th terms of $(1+x)^N$ have coefficients:

$$\binom{N}{4}, \quad \binom{N}{5}, \quad \binom{N}{6}$$

Step 2 — Impose the AP condition

$$2\binom{N}{5} = \binom{N}{4}+\binom{N}{6}$$

Dividing throughout by $\binom{N}{5}$:

$$2 = \frac{\binom{N}{4}}{\binom{N}{5}} + \frac{\binom{N}{6}}{\binom{N}{5}}$$

Step 3 — Use the standard consecutive-coefficient ratios

$$\frac{\binom{N}{4}}{\binom{N}{5}} = \frac{5}{N-4}, \qquad \frac{\binom{N}{6}}{\binom{N}{5}} = \frac{N-5}{6}$$

So:

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

Step 4 — Clear denominators and simplify

Multiplying through by $6(N-4)$:

$$12(N-4) = 30 + (N-5)(N-4)$$

$$12N-48 = 30 + N^2-9N+20$$

$$0 = N^2 – 21N + 98$$

Step 5 — Solve the quadratic

$$N = \frac{21 \pm \sqrt{441-392}}{2} = \frac{21\pm7}{2} = 14 \text{ or } 7$$

Since $N=n+4$: $n=10$ or $n=3$. The problem excludes $n=10$, so $n=3$ and $N=7$.

Step 6 — Find the largest coefficient of $(1+x)^7$

For an odd power like $7$, the two largest (and equal) binomial coefficients are the middle ones, $\binom{7}{3}$ and $\binom{7}{4}$:

$$\binom{7}{4} = \frac{7!}{4!\,3!} = \frac{7\times6\times5}{3\times2\times1} = 35$$

Answer

Common mistakes

• Forgetting that $(1+x)^7$ has two equal largest coefficients (at positions 4 and 5, i.e. $\binom{7}{3}=\binom{7}{4}$), since $7$ is odd — there’s no single unique middle term the way there would be for an even power.
• Not excluding $n=10$. The quadratic naturally produces both $n=3$ and $n=10$; missing the problem’s explicit exclusion leads to solving for the wrong expansion entirely.

Practise next

• For some $m\neq5$, the coefficients of the 5th, 6th and 7th terms of $(1+x)^{m+2}$ are in AP. Find $m$, and the largest coefficient of the resulting expansion.