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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2024 · Paper 1 · Mathematics · Sequences and Series

If arithmetic mean of two distinct positive real number a and b(a >...

If arithmetic mean of two distinct positive real number aa and b(a>b)b(a > b) be twice their geometric mean, then a:b=a: b=

Options

  1. A(2+3):(2−3)(2+\sqrt{3}):(2-\sqrt{3})
  2. B(2+5):(2−5)(2+\sqrt{5}):(2-\sqrt{5})
  3. C(2+2):(2−2)(2+\sqrt{2}):(2-\sqrt{2})
  4. DNone of these

Correct answer

A. (2+3):(2−3)(2+\sqrt{3}):(2-\sqrt{3})

Step-by-step solution

By the given condition, a+b2=2ab⇒a+b=4ab\frac{a+b}{2}=2 \sqrt{a b} \Rightarrow a+b=4 \sqrt{a b} Now, (a−b)2=(a+b)2−4ab=16ab−4ab=12ab∴a−b=12ab=23ab(a-b)^2=(a+b)^2-4 a b=16 a b-4 a b=12 a b\therefore a-b=\sqrt{12 a b}=2 \sqrt{3} \sqrt{a b}. (Taking + ve sign only as a>ba > b) ∴a+ba−b=4ab23ab=23\therefore \frac{a+b}{a-b}=\frac{4 \sqrt{a b}}{2 \sqrt{3} \sqrt{a b}}=\frac{2}{\sqrt{3}} By componendo and dividendo, 2a2b=2+32−3\frac{2 a}{2 b}=\frac{2+\sqrt{3}}{2-\sqrt{3}} or ab=2+32−3\frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}
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