Let a , b, c , d and p be any non zero distinct real numbers such that
(a² + b² + c²)p² – 2(ab + bc + cd)p + (b² + c² + d²) = 0. Then :

Solution

Maths sir · 4 steps

Regroup the left side into three perfect squares. Real squares adding to zero must each be zero.

1. Spot the squares. Pair the terms: a²p² - 2abp + b², then b²p² - 2bcp + c², then c²p² - 2cdp + d². Together they use every term of the equation exactly once.

   (ap - b)² + (bp - c)² + (cp - d)² = 0

2. Each square vanishes. All the numbers are real, so no square is negative. A sum of non-negative terms is zero only when each term is zero. That gives b = ap, c = bp and d = cp.
3. Read the pattern. Each number is p times the one before it, so b/a = c/b = d/c = p. That is a G.P. with common ratio p.
4. So. a, b, c, d are in G.P., which is option (b).

The idea to keep. When a real equation has squared terms and cross terms with a -2, try to write it as a sum of squares equal to zero. Then every square must be zero on its own.

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