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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)`

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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: \[ AM = \frac{a + b}{2} \] ...
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OBJECTIVE RD SHARMA ENGLISH-SEQUENCES AND SERIES-Chapter Test
  1. If the arithmetic mean of two numbers a and b,a>b>0, is five times the...

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  2. Let H(n)=1+(1)/(2)+(1)/(3)+ . . . . .+(1)/(n), then the sum to n terms...

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  3. Sum of the first n terms of the series 1/2+3/4+7/8+(15)/(16)+............

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  4. If A(1),A(2) are between two numbers, then (A(1)+A(2))/(H(1)+H(2)) is ...

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  5. If the (m+1)t h ,(n+1)t h ,a n d(r+1)t h terms of an A.P., are in G.P....

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  6. Given that n arithmetic means are inserted between two sets of numbers...

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  7. If a,b, and c are in G.P then a+b,2b and b+ c are in

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  8. If in a progression a1, a2, a3, e t cdot,(ar-a(r+1)) bears a constant...

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  9. If in an AP, t1 = log10 a, t(n+1) = log10 b and t(2n+1) = log10 c then...

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  10. Find the sum of the series: 1^2-2^2+3^2-4^2+.....-2008^2+2009^2.

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  11. If 4a^(2)+9b^(2)+16c^(2)=2(3ab+6bc+4ca)," where "a,b,c are non-zero nu...

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  12. If Sn denotes the sum of n terms of an A.P. whose common difference is...

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  13. The sides of a right angled triangle are in A.P., then they are in the...

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  14. Find the sum of all the 11 terms of an AP whose middle most term is 30...

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  15. The maximum sum of the series 20+19 1/3+18 2/3+ is 310 b. 300 c. 0320 ...

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  16. If three numbers are in G.P., then the numbers obtained by adding the ...

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  17. If p ,q ,r are in A.P., show that the pth, qth and rth terms of any G....

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  18. Let a,b,c be three positive prime number. The progrrssion in which sqr...

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  19. If 1/(b-a)+1/(b-c)=1/a+1/c , then (A). a ,b ,a n dc are in H.P. (B). a...

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  20. If three numbers are in H.P., then the numbers obtained by subtracting...

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  21. The first three of four given numbers are in G.P. and their last three...

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