A-Level Sequences and series: worked solution

Question

A geometric sequence {u_n} has first term a > 0 and common ratio 3/2. Define {v_n} by v_n = 9/u_n for every positive integer n. (a) Prove {v_n} is geometric too. [2] (b) If v_2 = u_2, determine a. [2]

(a) {u_n} is geometric with first term a and common ratio 3/2, so its nth term is u_n = a(3/2)^(n-1). To prove a sequence is geometric you must show the ratio of consecutive terms is a constant that does not depend on n. v_n / v_(n-1) = [9/u_n] divided by [9/u_(n-1)] = (9/u_n) x (u_(n-1)/9) = u_(n-1)/u_n. Since {u_n} is geometric with ratio 3/2, u_n = (3/2)u_(n-1), so u_(n-1)/u_n = 1/(3/2) = 2/3. This is the same value 2/3 for every n >= 2 and contains no n, so {v_n} is geometric, with first term v_1 = 9/a and common ratio 2/3. (b) Write down each of the two second terms. u_2 = a(3/2)^(2-1) = 3a/2. v_2 = 9/u_2 = 9 divided by (3a/2) = 9 x 2/(3a) = 6/a. Set them equal: 6/a = 3a/2. Multiply both sides by 2a (allowed, since a > 0 means a is not zero): 12 = 3a². Divide by 3: a² = 4, so a = 2 or a = -2. The question states a > 0, so a = -2 is rejected. Therefore a = 2.