Progression (Sequence and Series)Hard
Question
If a, b, c 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, d
Options
A.are in AP
B.are in GP
C.are in HP
D.satisfy ab = cd
Solution
Here, (a2 + b2 + c2 ) p2 - 2(ab + bc + cd) p + (b2 + c2 + d2) ≤ 0
⇒ (a2 p2 - 2abp + b2) + (b2 p2 - abcp + c2) + (c2 p2 - 2cdp + d2) ≤ 0
⇒ (ap - b)2 + (bp - c)2 + (cp - d)2 ≤ 0
(Since, sum of squares is never less than zero)
⇒ Each of the squares is zero
∴ (ap - b)2 = (bp - c)2 = (cp - d)2 = 0
⇒
∴ a, b, c, d are in GP.
⇒ (a2 p2 - 2abp + b2) + (b2 p2 - abcp + c2) + (c2 p2 - 2cdp + d2) ≤ 0
⇒ (ap - b)2 + (bp - c)2 + (cp - d)2 ≤ 0
(Since, sum of squares is never less than zero)
⇒ Each of the squares is zero
∴ (ap - b)2 = (bp - c)2 = (cp - d)2 = 0
⇒
∴ a, b, c, d are in GP.
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