JEE Main 2020 · 2 September · morning · Single correct · +4 / −1
The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie in :
- (a) [-3, ∞)
- (b) (-∝, 9]
- (c) (-∝, -9] ∪ [-3, ∞)
- (d) (-∝, -3] ∪ [9, ∞)
Answer and solution
Answer (d) (-∝, -3] ∪ [9, ∞)
Solution
Maths sir · 4 steps
Take the terms as a/r, a, ar: the product fixes a = 3, and then r + 1/r does the rest.
- Pick symmetric terms. Write the three terms as a/r, a, ar with r ≠ 0. Their product is a³, so a³ = 27 and a = 3.
S = 3/r + 3 + 3r = 3(1 + r + 1/r)
- Bound r + 1/r. For r > 0, AM–GM gives r + 1/r ≥ 2, with equality at r = 1. For r < 0, the same argument on -r gives r + 1/r ≤ -2, with equality at r = -1.
- Range of S. If r > 0, then S ≥ 3(1 + 2) = 9. If r < 0, then S ≤ 3(1 - 2) = -3. Both ends are reached: 3, 3, 3 gives 9 and -3, 3, -3 gives -3.
- So. S ∈ (-∞, -3] ∪ [9, ∞), which is option (d).
The idea to keep. When a G.P. question gives the product of three terms, take them as a/r, a, ar. Then r + 1/r is at least 2 or at most -2, so always split into r > 0 and r < 0.
Answer from the NTA final answer key. Solution by PrepWiser.