Problem 15 - Entrance Test

Let the three terms be a/r, a, ar (symmetric form for three terms in GP). Then sum: a/r + a + ar = 14 → a(1/r + 1 + r) = 14 ...(1). Sum of squares: a^2/r^2 + a^2 + a^2r^2 = 84 → a^2(1/r^2 + 1 + r^2) = 84 ...(2). Let x = r + 1/r. Then 1/r + 1 + r = x + 1. And 1/r^2 + 1 + r^2 = (r^2 + 1/r^2) + 1 = (x^2 - 2) + 1 = x^2 - 1. So equations become: a(x+1) = 14 ...(1') and a^2(x^2-1) = 84 ...(2'). From (1'): a = 14/(x+1). Substitute into (2'): [196/(x+1)^2] * (x^2-1) = 84. Note x^2-1 = (x-1)(x+1). So: 196(x-1)(x+1) / (x+1)^2 = 84 → 196(x-1)/(x+1) = 84 → (x-1)/(x+1) = 84/196 = 3/7. Cross multiply: 7(x-1) = 3(x+1) → 7x - 7 = 3x + 3 → 4x = 10 → x = 2.5 = 5/2. Now x = r + 1/r = 5/2. Multiply by r: r^2 + 1 = (5/2)r → 2r^2 - 5r + 2 = 0 → (2r-1)(r-2) = 0. So r = 1/2 or r = 2. Both are valid common ratios. Among the options, both 2 and 1/2 are present. Since both are valid, and the question likely expects the greater value, the answer is 2 (Option A). Note: r=1/2 is also correct, as the GP can be written in reverse order.