Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

If a, b, c, d and p are distinct real numbers such that (a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) ≤ 0, then a, b, c and d

If a, b, c, d and p are distinct real numbers such that (a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) ≤ 0, then a, b, c and d

(1) are in AP

(2) are in GP

(3) are in HP

(4) satisfy ab = cd

Solution:

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

(a2 p2 + b2 p2 + c2 p2 )- (2abp + 2bcp + cpd) + (b2 + c2 + d2) ≤ 0

(a2 p2 – 2abp + b2) + (b2p2 – 2bpc + c2) + (c2p2 – 2cdp + d2) ≤ 0

(ap -b)2 + (bp – c)2 + (cp – d)2 ≤ 0

Sum of squares cannot be negative.

So (ap -b) = 0

ap = b …(i)

(bp – c) = 0

bp = c …(ii)

cp – d = 0

cp = d …(iii)

From (i) and (ii)

b/a = c/b

=> b2 = ac

From (ii) and (iii)

c/b = d/c

c2 = bd

Hence a, b, c, d are in GP.

So option (2) is the answer.

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