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1987

JEE Main 1987 Mathematics question on Sequences and Series

If a,b,c,da, b, c, d and pp are distinct real numbers such that (a2+b2+c2) p2−2(ab+bc+cd) p+(b2+c2+d2)≤0(a^2 + b^2 + c^2)\, p^2 - 2(ab + bc + cd)\, p + (b^2 + c^2 + d^2) \leq 0, then a,b,c,da, b, c, d
  1. A.
    (a) are in AP
  2. B.
    (b) are in GP
  3. C.
    (c) are in HP
  4. D.
    (d) satisfy ab=cdab = cd
Show answer and solution

Correct option: B

Solution

The expression (a²+b²+c²)p² - 2(ab+bc+cd)p + (b²+c²+d²) ≤ 0 can be rewritten by completing the square. Notice it equals (ap-b)² + (bp-c)² + (cp-d)² ≤ 0. Since each squared term is non-negative, the sum can only be ≤ 0 if each term equals zero simultaneously. So ap = b, bp = c, cp = d, meaning b/a = c/b = d/c = p, which is exactly the condition for a, b, c, d to be in GP.

[1987]

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