Question

please answer this question I mark you as brainliest its urgent​

Answer 1

Explanation:

(a²+b²+c²)p²−2(ab+bc+cd)p+(b²+c²+d²)≤0

⇒(a²p²-2abp+b²)+(b²p²-2bcp+c²)+(c²p²−2cdp+d²)≤0

=>(ap−b)²+(bp−c)² +(cp−d)²≤0

Sum of square for real numbers can't be negative

Thus ap=b,bp=c,cp=d

⇒ b/a = c/b = d/c

⇒a,b,c,d are in G.P

Hope it helps you..

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