If n 13 , n 14 and n 15 are in arithmetic progression, then the positive integer value of ' n ' can be

Step-by-step solution

Given ^ n C₁₃ , ^ n C₁₄ , and ^ n C₁₅ are in arithmetic progression. 2 n 14 = n 13 + n 15 Dividing the equation by ^ n C₁₄ , we get: 2 = ^ n C₁₃ ^ n C₁₄ + ^ n C₁₅ ^ n C₁₄ Using the property ^ n C_ r ^ n C_ r-1 = n-r+1 r , we have: 2 = 14 n-13 + n-14 15 Multiplying by 15(n-13) on both sides: 30(n-13) = 14 15 + (n-14)(n-13) 30n - 390 = 210 + n^2 - 27n + 182 n^2 - 57n + 782 = 0 (n-23)(n-34) = 0 n = 23 or n = 34 From the given options, n = 34 . Answer: 34