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If the terms of G.P. sqrt(a-x),sqrt(x),s...

If the terms of G.P. `sqrt(a-x),sqrt(x),sqrt(a+x)`.. are all in integers where `agto` then the least composite odd integral value of `a` is

A

`9`

B

`15`

C

`45`

D

`49`

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The correct Answer is:
To solve the problem, we need to determine the least composite odd integral value of \( a \) such that the terms \( \sqrt{a - x}, \sqrt{x}, \sqrt{a + x} \) are all integers and \( a > 0 \). ### Step-by-Step Solution: 1. **Understanding the Condition**: The terms \( \sqrt{a - x}, \sqrt{x}, \sqrt{a + x} \) need to be in Arithmetic Progression (AP). For three terms \( A, B, C \) to be in AP, the condition is: \[ 2B = A + C \] Here, let \( A = \sqrt{a - x} \), \( B = \sqrt{x} \), and \( C = \sqrt{a + x} \). 2. **Setting Up the Equation**: From the AP condition, we have: \[ 2\sqrt{x} = \sqrt{a - x} + \sqrt{a + x} \] 3. **Squaring Both Sides**: To eliminate the square roots, we square both sides: \[ 4x = (\sqrt{a - x} + \sqrt{a + x})^2 \] Expanding the right-hand side: \[ 4x = (a - x) + (a + x) + 2\sqrt{(a - x)(a + x)} \] Simplifying gives: \[ 4x = 2a + 2\sqrt{(a - x)(a + x)} \] 4. **Rearranging the Equation**: Rearranging the equation, we get: \[ 2\sqrt{(a - x)(a + x)} = 4x - 2a \] Dividing by 2: \[ \sqrt{(a - x)(a + x)} = 2x - a \] 5. **Squaring Again**: Squaring both sides again gives: \[ (a - x)(a + x) = (2x - a)^2 \] Expanding both sides: \[ a^2 - x^2 = 4x^2 - 4ax + a^2 \] Simplifying leads to: \[ -x^2 = 4x^2 - 4ax \] Rearranging gives: \[ 5x^2 - 4ax = 0 \] 6. **Factoring**: Factoring out \( x \): \[ x(5x - 4a) = 0 \] This gives us two cases: \( x = 0 \) or \( 5x = 4a \). 7. **Finding \( x \)**: From \( 5x = 4a \), we have: \[ x = \frac{4a}{5} \] 8. **Substituting Back**: Now substituting \( x = \frac{4a}{5} \) back into the expressions for \( \sqrt{a - x}, \sqrt{x}, \sqrt{a + x} \): - \( \sqrt{x} = \sqrt{\frac{4a}{5}} \) - \( \sqrt{a - x} = \sqrt{a - \frac{4a}{5}} = \sqrt{\frac{a}{5}} \) - \( \sqrt{a + x} = \sqrt{a + \frac{4a}{5}} = \sqrt{\frac{9a}{5}} \) 9. **Ensuring Integer Values**: For these square roots to be integers, \( a \) must be of the form \( a = 5n^2 \) where \( n \) is a natural number. 10. **Finding Composite Odd Values**: - For \( n = 1 \): \( a = 5 \) (not composite) - For \( n = 2 \): \( a = 20 \) (composite but not odd) - For \( n = 3 \): \( a = 45 \) (composite and odd) Thus, the least composite odd integral value of \( a \) is: \[ \boxed{45} \]

To solve the problem, we need to determine the least composite odd integral value of \( a \) such that the terms \( \sqrt{a - x}, \sqrt{x}, \sqrt{a + x} \) are all integers and \( a > 0 \). ### Step-by-Step Solution: 1. **Understanding the Condition**: The terms \( \sqrt{a - x}, \sqrt{x}, \sqrt{a + x} \) need to be in Arithmetic Progression (AP). For three terms \( A, B, C \) to be in AP, the condition is: \[ 2B = A + C ...
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