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If a,b,c,d are such unequal real numbers...

If a,b,c,d are such unequal real numbers that `(a^(2)+b^(2)+c^(2))p^(2)-2 (ab+bc+cd) p+ (b^(2)+c^(2)+d^(2)) le 0` then a,b,c, d are in -

A

are in AP

B

are in GP

C

are in HP

D

satisfy ab = cd

Text Solution

Verified by Experts

The correct Answer is:
b
Here ,`(a^(2)+b^(2)+c^(2))p^(2)-2 (ab+bc+cd)p`
`+(b^(2)+c^(2)+d^(2)) le 0`
` (a^(2)p^(2)-2abp+b^(2)) +(b^(2)p^(2)-2bcp +c^(2))`
` +(c^(2)p^(2)-2cdp+d^(2))le 0`
` rArr (ap-b)^(2) +(bp-c)^(2)+(cp-d)^(2)le 0`
[ since, sum of squares is never less than zero ] ,
Each of the square is zero .
` :. (ap-b)^(2) = (bp -c)^(2) = (cp -d)^(2)=0 `
` rArr p = b/a =c/b = d/c`
So,a,b,c and d are in GP .
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