Question:

<p><strong>Step 1:</strong> If <span class="math">a, b, c</span> are in AP, then <span class="math">2b = a + c</span>.</p><p>Therefore: <span class="math">2 \cdot \binom{n}{5} = \binom{n}{4} + \binom{n}{6}</span></p><p><strong>Step 2:</strong> Substitute the formula <span class="math">\binom{n}{r} = \frac{n!}{r!(n-r)!}</span>:</p><p><span class="math">2 \cdot \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!}</span></p><p><strong>Step 3:</strong> Divide through by <span class="math">\frac{n!}{4!(n-4)!}</span> and simplify:</p><p><span class="math">\frac{2}{5(n-5)} = \frac{1}{(n-4)(n-5)} + \frac{1}{30}</span></p><p><strong>Step 4:</strong> Combine fractions:</p><p><span class="math">\frac{2}{5(n-5)} = \frac{30 + (n-4)(n-5)}{30(n-4)(n-5)}</span></p><p><strong>Step 5:</strong> Cross-multiply and simplify:</p><p><span class="math">12(n-4) = 30 + n^2 - 9n + 20</span></p><p><span class="math">n^2 - 21n + 98 = 0</span></p><p><strong>Step 6:</strong> Factor:</p><p><span class="math">n^2 - 14n - 7n + 98 = 0</span></p><p><span class="math">n(n-14) - 7(n-14) = 0</span></p><p><span class="math">(n-7)(n-14) = 0</span></p><p><strong>∴ Answer: n = 7 or n = 14</strong></p>