Given the sequences u0=1, un+1=(1)/(3)(un+2vn), v0=12, vn+1=(1)/(4)(un+3vn), and wn=vn-un, show that (wn) is a geometric sequence with positive terms.

Here's the solution for question 10. Given the sequences: u_0 = 1 u_n+1 = (1)/(3)(u_n + 2v_n) v_0 = 12 v_n+1 = (1)/(4)(u_n + 3v_n) w_n = v_n - u_n i) Show that (w_n) is a geometric sequence with positive terms. Step 1: Express w_n+1 in terms of u_n and v_n. w_n+1 = v_n+1 - u_n+1 Substitute the given recurrence relations for u_n+1 and v_n+1: w_n+1 = (1)/(4)(u_n + 3v_n) - (1)/(3)(u_n + 2v_n) Step 2: Simplify the expression for w_n+1. Find a common denominator, which is 12: w_n+1 = (3(u_n + 3v_n) - 4(u_n + 2v_n))/(12) w_n+1 = (3u_n + 9v_n - 4u_n - 8v_n)/(12) w_n+1 = (-u_n + v_n)/(12) w_n+1 = (v_n - u_n)/(12) Step 3: Relate w_n+1 to w_n. Since w_n = v_n - u_n, we have: w_n+1 = (1)/(12) w_n This shows that (w_n) is a geometric sequence with a common ratio r = (1)/(12). Step 4: Check if the terms are positive. Calculate the first term w_0: