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

Welcome back RÉSPÊÇT📿☝️.......🙏🙏🔐 — been a while! Let's pick up where you left off. Step 1: Identify the coefficients of the 5th, 6th, and 7th terms. The general term in the binomial expansion of (1+x)^n is T_r+1 = nr x^r. The coefficient of the (r+1)-th term is nr. • The coefficient of the 5th term (for r=4) is C_5 = n4. • The coefficient of the 6th term (for r=5) is C_6 = n5. • The coefficient of the 7th term (for r=6) is C_7 = n6. Step 2: Apply the condition for an arithmetic progression (A.P.). 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 n5 = n4 + n6 Step 3: Express the binomial coefficients using the identity nk = (n-k+1)/(k) nk-1. We can write n4 and n6 in terms of n5: • From n5 = (n-4)/(5) n4, we get n4 = (5)/(n-4) n5. • From n6 = (n-5)/(6) n5. Step 4: Substitute these expressions into the A.P. equation. 2 n5 = (5)/(n-4) n5 + (n-5)/(6) n5 Since n5 cannot be zero (as n 5 for the 6th term to exist), we can divide both sides by n5: 2 = (5)/(n-4) + (n-5)/(6) Step 5: Solve the equation for n. Multiply the entire equation by the common denominator 6(n-4) to eliminate fractions: 2 × 6(n-4) = 5 × 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 Step 6: Factor the quadratic equation. We need two numbers that multiply to 98 and add up to -21. These numbers are -7 and -14. (n-7)(n-14) = 0 Step 7: Determine the possible values of n. Setting each factor to zero gives the possible values for n: n-7 = 0 n=7 n-14 = 0 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 7, 14. Send me the next one 📸