If and be three distinct real numbers in GP and , then cannot be
- A.4
- B.2
- C.
- D.
Show answer and solution
Correct option: B
Solution
Let b = ar and c = ar², so a + b + c = a(1 + r + r²) and xb = xar.
Thus x = (1 + r + r²)/r = r + 1 + 1/r.
Let t = r + 1/r. Since a, b, c are distinct reals in GP, r ≠ 0, 1, -1, so t = r + 1/r can take any value except those in (-2, 2) — specifically t ∈ (-∞, -2) ∪ (2, ∞).
Therefore x = t + 1 ∈ (-∞, -1) ∪ (3, ∞).
So x cannot equal 2 (since 2 lies in [-1, 3], outside the achievable range). Options like 4, -2, -3 all lie in the valid range, but x = 2 is impossible.
Thus x = (1 + r + r²)/r = r + 1 + 1/r.
Let t = r + 1/r. Since a, b, c are distinct reals in GP, r ≠ 0, 1, -1, so t = r + 1/r can take any value except those in (-2, 2) — specifically t ∈ (-∞, -2) ∪ (2, ∞).
Therefore x = t + 1 ∈ (-∞, -1) ∪ (3, ∞).
So x cannot equal 2 (since 2 lies in [-1, 3], outside the achievable range). Options like 4, -2, -3 all lie in the valid range, but x = 2 is impossible.
[JEE Main 2019 · 9 Jan · Shift I]