Home
Class 11
MATHS
If the arithmetic mean of two numbers a ...

If the arithmetic mean of two numbers a and b,`a>b>0`, is five times their geometric mean, then `(a+b)/(a-b)` is equal to:

A

`2+sqrt(3):2-sqrt(3)`

B

`7+4sqrt(3):7-4sqrt(3)`

C

`2:7+4sqrt(3)`

D

`2:sqrt(3)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we start with the given conditions and derive the required expression. ### Step 1: Understand the given conditions We know that: - The arithmetic mean (AM) of two numbers \( a \) and \( b \) is given by: \[ AM = \frac{a + b}{2} \] - The geometric mean (GM) of the same two numbers is given by: \[ GM = \sqrt{ab} \] ### Step 2: Set up the equation based on the problem statement According to the problem, the arithmetic mean is five times the geometric mean: \[ \frac{a + b}{2} = 5 \sqrt{ab} \] ### Step 3: Multiply both sides by 2 To eliminate the fraction, we multiply both sides by 2: \[ a + b = 10 \sqrt{ab} \] ### Step 4: Square both sides Next, we square both sides to eliminate the square root: \[ (a + b)^2 = (10 \sqrt{ab})^2 \] This simplifies to: \[ a^2 + 2ab + b^2 = 100ab \] ### Step 5: Rearrange the equation Rearranging the equation gives: \[ a^2 + b^2 + 2ab - 100ab = 0 \] \[ a^2 + b^2 - 98ab = 0 \] ### Step 6: Use the identity for \( a - b \) We can use the identity: \[ a^2 - b^2 = (a + b)(a - b) \] From the earlier step, we know: \[ a^2 + b^2 = (a + b)^2 - 2ab \] Substituting this in gives: \[ (a + b)^2 - 2ab - 98ab = 0 \] Thus: \[ (a + b)^2 - 100ab = 0 \] ### Step 7: Express \( a - b \) Now, we use the identity: \[ a - b = \sqrt{(a + b)^2 - 4ab} \] Substituting \( (a + b)^2 = 100ab \): \[ a - b = \sqrt{100ab - 4ab} = \sqrt{96ab} \] ### Step 8: Find \( \frac{a + b}{a - b} \) Now we need to find: \[ \frac{a + b}{a - b} \] Substituting the expressions we have: \[ \frac{a + b}{a - b} = \frac{10 \sqrt{ab}}{\sqrt{96ab}} = \frac{10}{\sqrt{96}} = \frac{10}{4\sqrt{6}} = \frac{5}{2\sqrt{6}} \] ### Step 9: Rationalize the denominator To rationalize the denominator: \[ \frac{5}{2\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{5\sqrt{6}}{12} \] ### Final Answer Thus, the value of \( \frac{a + b}{a - b} \) is: \[ \frac{5\sqrt{6}}{12} \]

To solve the problem step by step, we start with the given conditions and derive the required expression. ### Step 1: Understand the given conditions We know that: - The arithmetic mean (AM) of two numbers \( a \) and \( b \) is given by: \[ AM = \frac{a + b}{2} \] ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • SEQUENCES AND SERIES

    OBJECTIVE RD SHARMA ENGLISH|Exercise Section I - Solved Mcqs|80 Videos
  • SEQUENCES AND SERIES

    OBJECTIVE RD SHARMA ENGLISH|Exercise Section II - Assertion Reason Type|13 Videos
  • QUADRATIC EXPRESSIONS AND EQUATIONS

    OBJECTIVE RD SHARMA ENGLISH|Exercise Chapter Test|50 Videos
  • SETS

    OBJECTIVE RD SHARMA ENGLISH|Exercise Chapter Test|30 Videos

Similar Questions

Explore conceptually related problems

If the arithmetic means of two positive number a and b (a gt b ) is twice their geometric mean, then find the ratio a: b

The arithmetic mean of two positive numbers a and b exceeds their geometric mean by 2 and the harmonic mean is one - fifth of the greater of a and b, such that alpha=a+b and beta=|a-b| , then the value of alpha+beta^(2) is equal to

Knowledge Check

  • {:("Column A","The arithmetic mean (average) of the number a and b is 17. The geometric mean of the numbers a and b is 8. The geometric mean of two numbers is defined to be the square root of their product","Column B"),(a,,b):}

    A
    If column A is larger
    B
    If column B is larger
    C
    If the columns are equal
    D
    If there is not enough information to decide
  • Similar Questions

    Explore conceptually related problems

    The arithmetic mean of two positive numbers a and b exceeds their geometric mean by (3)/(2) and the geometric mean exceeds their harmonic mean by (6)/(5) . If a+b=alpha and |a-b|=beta, then the value of (10beta)/(alpha) is equal to

    If arithmetic mean of two positive numbers is A, their geometric mean is G and harmonic mean H, then H is equal to

    The arithmetic mean between two numbers is A and the geometric mean is G. Then these numbers are:

    Find the arithmetic mean of first five prime numbers.

    The arithmetic mean between a and b is twice the geometric mean between a and b. Prove that : (a)/(b) = 7+ 4sqrt3 or 7- 4sqrt3

    If the arithmetic mean of the numbers x_1, x_2, x_3, ..., x_n is barx, then the arithmetic mean of the numbers ax_1 +b, ax_2 +b, ax_3 +b, ....,ax_n +b, where a, b are two constants, would be

    The arithmetic mean of two numbers is 6 and their geometric mean G and harmonic mean H satisfy the relation G^2+3H=48 .Find the two numbers.