If three successive terms of a G.P. with common ratio r(r>1) are the lengths of the sides of a triangle and [r] denotes the greatest integer less than or equal to r, then 3[r]+[-r] is equal to _____________.

Solution

Maths sir · 4 steps

Take the sides as a, ar, ar²: the two shorter sides must add up to more than the longest.

1. Set up the sides. Let the sides be a, ar, ar² with a > 0. Since r > 1, the longest side is ar². The other two triangle inequalities hold automatically, so only one condition matters.

   a + ar > ar² ⇒ r² - r - 1 < 0

2. Solve for r. The roots of r² - r - 1 = 0 are (1 ± √5)/2. The quadratic is negative between them. With r > 1 this gives 1 < r < (1+√5)/2 ≈ 1.618.
3. Find the floors. For 1 < r < 1.618 we get [r] = 1. Then -1.618 < -r < -1, and the greatest integer not exceeding -r is -2, so [-r] = -2.
4. So. 3[r] + [-r] = 3(1) + (-2) = 1.

The common slip: 2. Taking [-r] = -1 by just copying the sign of [r] gives 3 - 1 = 2. For a non-integer r, [-r] = -[r] - 1 = -2.

The idea to keep. For G.P. sides a, ar, ar², the triangle condition traps r between (√5-1)/2 and (√5+1)/2. And for any non-integer x, [-x] = -[x] - 1.

Answer from the NTA final answer key. Solution by PrepWiser.