# If a,b,c are in GP and a + x , b + x, c + x are in HP. Find x where a,b,c are distinct numbers?

Aug 6, 2017

$x = b .$

#### Explanation:

As $a , b$ and $c$ are in GP, we have ${b}^{2} = a c$.

Further as $a , b , c$ are distinct numbers, $a \ne b \ne c \ne a$

If $a + x$, $b + x$ and $c + x$ are in HP, then

$\frac{1}{a + x} , \frac{1}{b + x}$ and $\frac{1}{c + x}$ are in AP and therefore

$\frac{1}{c + x} - \frac{1}{b + x} = \frac{1}{b + x} - \frac{1}{a + x}$

i.e. $\frac{b + x - c - x}{\left(c + x\right) \left(b + x\right)} = \frac{a + x - b - x}{\left(b + x\right) \left(a + x\right)}$

i.e. $\frac{b - c}{b c + b x + c x + {x}^{2}} = \frac{a - b}{b a + b x + a x + {x}^{2}}$

or $\left(b - c\right) \left(b a + b x + a x + {x}^{2}\right) = \left(a - b\right) \left(b c + b x + c x + {x}^{2}\right)$

or ${b}^{2} a + {b}^{2} x + a b x + b {x}^{2} - a b c - b c x - a c x - c {x}^{2} = a b c + a b x + a c x + a {x}^{2} - {b}^{2} c - {b}^{2} x - b c x - b {x}^{2}$

or ${b}^{2} a + 2 {b}^{2} x + 2 b {x}^{2} - 2 a b c - 2 a c x - c {x}^{2} - a {x}^{2} + {b}^{2} c = 0$

or ${x}^{2} \left(2 b - c - a\right) + x \left(2 {b}^{2} - 2 a c\right) + {b}^{2} a - 2 a b c + {b}^{2} c = 0$

As ${b}^{2} = a c$, this becomes ${x}^{2} \left(2 b - c - a\right) + {b}^{2} \left(a - 2 b + c\right) = 0 ,$

or ${x}^{2} \left(2 b - c - a\right) - {b}^{2} \left(2 b - c - a\right) = 0 ,$

$\Rightarrow \left(2 b - c - a\right) \left({x}^{2} - {b}^{2}\right) = 0.$

 rArr {(b-c)+(b-a)}}(x-b)(x+b)=0.

Since, $a \ne b \ne c , \left(b - c\right) \ne 0 , \left(b - a\right) \ne 0 , \therefore \left(b - c\right) + \left(c - a\right) \ne 0.$

Also, $x + b \ne 0.$

Clearly, $x = b .$

Aug 6, 2017