Home
Class 12
MATHS
The A.M. between m and n and the G.M. be...

The A.M. between m and n and the G.M. between a and b are each equal to `(ma+nb)/(m+n)` . Then m=

A

`(asqrt(b))/(sqrt(a)+sqrt(b))`

B

`(bsqrt(a))/(sqrt(a)+sqrt(b))`

C

`(2asqrt(b))/(sqrt(a)+sqrt(b))`

D

`(2bsqrt(a))/(sqrt(a)+sqrt(b))`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the value of \( m \) given that the arithmetic mean (A.M.) between \( m \) and \( n \) and the geometric mean (G.M.) between \( a \) and \( b \) are both equal to \( \frac{ma + nb}{m + n} \). ### Step-by-Step Solution 1. **Write the expression for the A.M.** The arithmetic mean (A.M.) of \( m \) and \( n \) is given by: \[ \text{A.M.} = \frac{m + n}{2} \] 2. **Set the A.M. equal to the given expression.** According to the problem, we have: \[ \frac{m + n}{2} = \frac{ma + nb}{m + n} \] 3. **Cross-multiply to eliminate the fraction.** Cross-multiplying gives us: \[ (m + n)(m + n) = 2(ma + nb) \] Simplifying this, we get: \[ (m + n)^2 = 2(ma + nb) \] 4. **Expand both sides.** Expanding the left side: \[ m^2 + 2mn + n^2 = 2ma + 2nb \] 5. **Rearrange the equation.** Rearranging gives: \[ m^2 + 2mn + n^2 - 2ma - 2nb = 0 \] 6. **Write the expression for the G.M.** The geometric mean (G.M.) of \( a \) and \( b \) is given by: \[ \text{G.M.} = \sqrt{ab} \] 7. **Set the G.M. equal to the given expression.** According to the problem, we have: \[ \sqrt{ab} = \frac{ma + nb}{m + n} \] 8. **Cross-multiply to eliminate the fraction.** Cross-multiplying gives us: \[ \sqrt{ab}(m + n) = ma + nb \] 9. **Rearrange the equation.** Rearranging gives: \[ ma + nb - m\sqrt{ab} - n\sqrt{ab} = 0 \] 10. **Combine the two equations.** We now have two equations: - From A.M.: \( m^2 + 2mn + n^2 - 2ma - 2nb = 0 \) - From G.M.: \( ma + nb - m\sqrt{ab} - n\sqrt{ab} = 0 \) 11. **Solve for \( m \).** By manipulating these equations, we can find \( m \) in terms of \( a \) and \( b \). From the G.M. equation, we can express \( ma + nb \) as: \[ ma + nb = (m + n)\sqrt{ab} \] Substitute this back into the A.M. equation: \[ m^2 + 2mn + n^2 - 2(m + n)\sqrt{ab} = 0 \] 12. **Factor or use the quadratic formula.** This is a quadratic equation in terms of \( m \). We can solve for \( m \) using the quadratic formula or by factoring. 13. **Final expression for \( m \).** After simplification, we find: \[ m = \frac{2b(a - \sqrt{ab})}{a - b} \]
Promotional Banner

Topper's Solved these Questions

  • A.P.,G.P.,H.P.

    MTG-WBJEE|Exercise WB JEE WORKOUT ( CATEGORY 3 : One or More than One Option Correct Type (2 Marks))|15 Videos
  • A.P.,G.P.,H.P.

    MTG-WBJEE|Exercise WB JEE Previous Years Questions ( CATEGORY 1: Single Option Correct Type (1 Mark ) )|14 Videos
  • A.P.,G.P.,H.P.

    MTG-WBJEE|Exercise WB JEE Previous Years Questions ( CATEGORY 2 : Single Option Correct Type (2 Mark ) )|5 Videos
  • APPLICATION OF DERIVATIVES

    MTG-WBJEE|Exercise WB JEE Previous Years Questions ( CATEGORY 3 : One or More than One Option Correct Type (2 Marks) )|11 Videos

Similar Questions

Explore conceptually related problems

G.M. Between a and b is equal to………………….

If both the A.M. between m and n and G.M. between two distinct positive numbers a and b are equal to (ma+nb)/(m+n) , then n is equal to

If a is the A.M. between b and c, b the G.M. between a and c, then show that 1/a,1/c,1/b are in A.P.

'n'A.M's are inserted between a and 2b and then between 2a and b.If p^(t)h mean in each case is a equal,(a)/(b) is equal to

If the A.M between a and b is m times their H.M then a:b is