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  • Stuck here, help me understand: - Sequence and series - JEE Main-3

If the arithmetic mean of two numbers a and b, a > b > 0, is five times their geometric

mean, then \frac{a+b}{a-b}     is equal to:

  • Option 1)

    \frac{\sqrt{6}}{2}

  • Option 2)

    \frac{3\sqrt{2}}{4}

  • Option 3)

    \frac{7\sqrt{3}}{12}

  • Option 4)

    \frac{5\sqrt{6}}{12}

 

Answers (3)

best_answer

As learned in

Arithmetic mean of two numbers (AM) -

A=\frac{a+b}{2}

- wherein

It is to be noted that the sequence a, A, b, is in AP where, a and b are the two numbers.

 

 and

Geometric mean of two numbers (GM) -

GM= \sqrt{ab}

- wherein

It is to be noted that a,G,b are in GP and a,b are two non - zero numbers.

 

 If \frac{a + b}{2} = 5 \sqrt{ab}

a + b = 10 \sqrt{ab}

\left ( a - b \right )^{2} = \left ( a + b \right )^{2} - 4ab

=> \left ( 10\sqrt{ab}\right )^{2} - 4ab = 100ab - 4ab = 96ab

a - b = 4\sqrt{6}\sqrt{ab}

Therefore:

\frac{a + b}{a - b} = \frac{10\sqrt{ab}}{4\sqrt{6}\sqrt{ab}} = 5 \frac{\sqrt{6}}{12}

 

 

 


Option 1)

\frac{\sqrt{6}}{2}

Option 2)

\frac{3\sqrt{2}}{4}

Option 3)

\frac{7\sqrt{3}}{12}

Option 4)

\frac{5\sqrt{6}}{12}

Posted by

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