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Question: If the terms of the AP \(\sqrt{a-x},\sqrt{x},\sqrt{a+x},.........\) are all integers where a > x > 0...

If the terms of the AP a−x,x,a+x,.........\sqrt{a-x},\sqrt{x},\sqrt{a+x},......... are all integers where a > x > 0, then find the least composite odd integral value of a.

Explanation

Solution

Hint: Hence the given three terms are in AP therefore, we will use the relation in AP that if there are three terms then the middle term is the mean of the other two terms. And by using this we will get a relation between a and x, then we will find the value of a in terms of x. And after that we will put the value of x such that we can get the value of a as the least composite odd integer.

Complete step-by-step answer:

Using the fact that if there are three terms in AP then the middle term is the mean of the other two terms.
Hence, from this we get,
x=a−x+a+x2 2x=a−x+a+x \begin{aligned} & \sqrt{x}=\dfrac{\sqrt{a-x}+\sqrt{a+x}}{2} \\\ & 2\sqrt{x}=\sqrt{a-x}+\sqrt{a+x} \\\ \end{aligned}
Now taking square on both the sides we get,
(2x)2=(a−x+a+x)2{{\left( 2\sqrt{x} \right)}^{2}}={{\left( \sqrt{a-x}+\sqrt{a+x} \right)}^{2}}
Now we will use the formula (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab to expand.
Therefore, we get
4x=(a+x)2+(a−x)2+2(a−x)(a+x)4x={{\left( \sqrt{a+x} \right)}^{2}}+{{\left( \sqrt{a-x} \right)}^{2}}+2\left( \sqrt{a-x} \right)\left( \sqrt{a+x} \right)
Now using the formula (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} we get,
4x=a+x+a−x+2a2−x2 4x=2a+2a2−x2 2x=a+a2−x2 2x−a=a2−x2 \begin{aligned} & 4x=a+x+a-x+2\sqrt{{{a}^{2}}-{{x}^{2}}} \\\ & 4x=2a+2\sqrt{{{a}^{2}}-{{x}^{2}}} \\\ & 2x=a+\sqrt{{{a}^{2}}-{{x}^{2}}} \\\ & 2x-a=\sqrt{{{a}^{2}}-{{x}^{2}}} \\\ \end{aligned}
Now taking square on both sides we get,
(2x−a)2=(a2−x2)2{{\left( 2x-a \right)}^{2}}={{\left( \sqrt{{{a}^{2}}-{{x}^{2}}} \right)}^{2}}
Now we will use the formula (a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab to expand.
Therefore, we get
(2x)2+a2−4ax=a2−x2 4x2+x2−4ax=a2−a2 5x2−4ax=0 x(5x−4a)=0 \begin{aligned} & {{\left( 2x \right)}^{2}}+{{a}^{2}}-4ax={{a}^{2}}-{{x}^{2}} \\\ & 4{{x}^{2}}+{{x}^{2}}-4ax={{a}^{2}}-{{a}^{2}} \\\ & 5{{x}^{2}}-4ax=0 \\\ & x\left( 5x-4a \right)=0 \\\ \end{aligned}
Now we have got two solution for this equation, x = 0 and 5x – 4a = 0
Now it is given in the question that x>0, so x can’t be equal to 0.
Hence, we get
5x=4a a=5x4 \begin{aligned} & 5x=4a \\\ & a=\dfrac{5x}{4} \\\ \end{aligned}
Now ‘a’ must be the least composite odd integer.
To satisfy the above condition x must be a multiple of 4 and for odd and composite it can be 4×34\times 3
Hence, we get
a=5×4×34=15a=\dfrac{5\times 4\times 3}{4}=15
Hence, the value of a = 15.

Note: As per the information given in the question the fact that we have used for AP is very important and must be kept in mind. One can prove the fact by taking any variable AP and see that it is true. All the algebraic formulas that we have used must be kept in mind. One must use the right simplifications by taking the terms to LHS or RHS and trying to remove the square root by squaring both sides. This is very helpful in simplifying terms having square roots.