# If the coefficients of 5th, 6th and 7th terms in the expansion of (1+x)^n are in A.P., then the value of n is A. 7 only B. 14 only C. 7 or 14 D. None of these

## Question

If the coefficients of $5^{\text{th}}, 6^{\text{th}}$ and $7^{\text{th}}$ terms in the expansion of $(1+x)^{\mathrm{n}}$ are in A.P., then the value of $n$ is

A. 7 only

B. 14 only

C. 7 or 14

D. None of these

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## Answer

The correct choice is **C. 7 or 14**.

To solve this, note that the coefficients of the terms in the binomial expansion $(1+x)^n$ for term $T_{r+1}$ are given by the binomial coefficient $\binom{n}{r}$. So, the coefficients for $5^{\text{th}}$, $6^{\text{th}}$ and $7^{\text{th}}$ terms are $\binom{n}{4}$, $\binom{n}{5}$, and $\binom{n}{6}$ respectively.

Given that these coefficients are in arithmetic progression (AP): $$ 2\binom{n}{5} = \binom{n}{4} + \binom{n}{6} $$

Using the property of binomial coefficients: $$ \binom{n}{4} = \frac{n!}{4!(n-4)!}, \quad \binom{n}{5} = \frac{n!}{5!(n-5)!}, \quad \binom{n}{6} = \frac{n!}{6!(n-6)!} $$

Simplifying and equating (by multiplying through by the appropriate factorial terms to clear the denominators): $$ 2\left(\frac{n!}{5!(n-5)!}\right) = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} $$

Rewriting further: $$ 2\left(\frac{1}{(n-5)5}\right) = \left(\frac{1}{(n-4)4}\right) + \left(\frac{1}{6 \times 5}\right) $$

This equals: $$ \frac{2}{(n-5)5} = \frac{1}{4(n-4)} + \frac{1}{30} $$

Multiplying through by $120(n-5)(n-4)$ gives: $$ 24(n-4) = 30(n-5) + 4(n-4)(n-5) $$

Expanding and simplifying: $$ 24n - 96 = 30n - 150 + 4(n^2 - 9n + 20) $$ $$ 4n^2 - 21n + 98 = 0 $$

This quadratic equation can be solved to find $n$, determining: $$ n=7 \text{ or } n=14 $$

Thus, the values of $n$ that meet the conditions of the problem are **7 or 14**.

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