The 5th, 6th, and 7th terms in the expansion of (1+x)ⁿ are in arithmetic progression. Find the value of n.

The general term in the binomial expansion of $(1+x)^n$ is $T_{r+1} = \binom{n}{r} x^r$. The coefficient of the $(r+1)$-th term is $\binom{n}{r}$.

• The coefficient of the 5th term (for $r=4$) is $C_5 = \binom{n}{4}$.
• The coefficient of the 6th term (for $r=5$) is $C_6 = \binom{n}{5}$.
• The coefficient of the 7th term (for $r=6$) is $C_7 = \binom{n}{6}$.

If three terms $a, b, c$ are in an A.P., then $2b = a+c$.
For the coefficients $C_5, C_6, C_7$:

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

We can write $\binom{n}{4}$ and $\binom{n}{6}$ in terms of $\binom{n}{5}$:

• From $\binom{n}{5} = \frac{n-4}{5} \binom{n}{4}$, we get $\binom{n}{4} = \frac{5}{n-4} \binom{n}{5}$.
• From $\binom{n}{6} = \frac{n-5}{6} \binom{n}{5}$.

equation.

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

Since $\binom{n}{5}$ cannot be zero (as $n \ge 5$ for the 6th term to exist), we can divide both sides by $\binom{n}{5}$:

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

Multiply the entire equation by the common denominator $6(n-4)$ to eliminate fractions:

$$2 \times 6(n-4) = 5 \times 6 + (n-5)(n-4)$$ $$12(n-4) = 30 + (n^2 - 4n - 5n + 20)$$ $$12n - 48 = 30 + n^2 - 9n + 20$$ $$12n - 48 = n^2 - 9n + 50$$

Rearrange the terms to form a quadratic equation:

$$n^2 - 9n - 12n + 50 + 48 = 0$$ $$n^2 - 21n + 98 = 0$$

We need two numbers that multiply to 98 and add up to -21. These numbers are -7 and -14.

$$(n-7)(n-14) = 0$$

Setting each factor to zero gives the possible values for $n$:

$$n-7 = 0 \implies n=7$$ $$n-14 = 0 \implies n=14$$

Both values are valid since $n$ must be at least 6 for the 7th term to exist in the expansion.

The possible values of $n$ are $\boxed{\text{7, 14}}$.